232 + Interview Questions in QUANTITATIVE APTITUDE in TRIGONOMETRY Page 4 InterviewSolution

151.	Find the Minimum value of 9 cos 2x + 2 sec 2x1). 3√22). 2√33). 6√34). 6√2
Answer» Solution: According to AM-GM inequality, (a + B)/2 ≥ √ab Here a = 9 cos²x and b = 2 sec²x Note: AM- GM inequality can be used ANYWHERE keeping in mind that a and b are positive. So, (9 cos²x + 2 sec²x)/2≥√(9cos²x)(2sec²x) (9 cos²x + 2 sec²x)/2≥√(9/sec²x)(2sec²x) (9 cos²x + 2 sec²x)/2≥√(9)(2) (9 cos²x + 2 sec²x)/2≥√18 (9 cos²x + 2 sec²x)≥2√18 (9 cos²x + 2 sec²x)≥2×3√2 (9 cos²x + 2 sec²x)≥6√2 So, the minimum value is 6√2. The correct OPTION is 4). 6√2

151.

Find the Minimum value of 9 cos 2x + 2 sec 2x1). 3√22). 2√33). 6√34). 6√2

Answer»

Solution:

According to AM-GM inequality,

(a + B)/2 ≥ √ab

Here a = 9 cos²x and b = 2 sec²x

Note: AM- GM inequality can be used ANYWHERE keeping in mind that a and b are positive.

So, (9 cos²x + 2 sec²x)/2≥√(9cos²x)(2sec²x)

(9 cos²x + 2 sec²x)/2≥√(9/sec²x)(2sec²x)

(9 cos²x + 2 sec²x)/2≥√(9)(2)

(9 cos²x + 2 sec²x)/2≥√18

(9 cos²x + 2 sec²x)≥2√18

(9 cos²x + 2 sec²x)≥2×3√2

(9 cos²x + 2 sec²x)≥6√2

So, the minimum value is 6√2.

The correct OPTION is 4). 6√2

Discussion

152.	1). 12). π/23). 5/24). None of these
Answer» We have, sin θ = 21/29 $(\Rightarrow {\RM{cos\theta }} = \sqrt {1 - {{\sin }^2}{\rm{\theta }}} = {\rm{\;}}\sqrt {1 - {{\left( {\FRAC{{21}}{{29}}} \RIGHT)}^2}} = {\rm{\;}}\sqrt {({{29}^2} - {{21}^2})/\left( {{{29}^2}} \right)} = \sqrt {\frac{{8 \times 50}}{{{{29}^2}}}} = {\rm{\;}}\frac{{20}}{{29}}{\rm{\;}})$ [We take only positive value of cos because θ lies between 0 and π/2] Now, $(\sec {\rm{\theta \;}} + {\rm{\;tan\theta }} = {\rm{\;}}\frac{1}{{{\rm{cos\theta }}}}{\rm{\;}} + {\rm{\;}}\frac{{{\rm{sin\theta }}}}{{{\rm{cos\theta }}}} = {\rm{\;}}\frac{{1{\rm{\;}} + {\rm{\;sin\theta }}}}{{{\rm{cos\theta }}}} = {\rm{\;}}\frac{{1{\rm{\;}} + {\rm{\;}}\frac{{21}}{{29}}}}{{\frac{{20}}{{29}}}} = {\rm{\;}}\frac{{50}}{{20}} = {\rm{\;}}\frac{5}{2})$

152.

1). 12). π/23). 5/24). None of these

Answer»

We have, sin θ = 21/29

$(\Rightarrow {\RM{cos\theta }} = \sqrt {1 - {{\sin }^2}{\rm{\theta }}} = {\rm{\;}}\sqrt {1 - {{\left( {\FRAC{{21}}{{29}}} \RIGHT)}^2}} = {\rm{\;}}\sqrt {({{29}^2} - {{21}^2})/\left( {{{29}^2}} \right)} = \sqrt {\frac{{8 \times 50}}{{{{29}^2}}}} = {\rm{\;}}\frac{{20}}{{29}}{\rm{\;}})$

[We take only positive value of cos because θ lies between 0 and π/2]

Now, $(\sec {\rm{\theta \;}} + {\rm{\;tan\theta }} = {\rm{\;}}\frac{1}{{{\rm{cos\theta }}}}{\rm{\;}} + {\rm{\;}}\frac{{{\rm{sin\theta }}}}{{{\rm{cos\theta }}}} = {\rm{\;}}\frac{{1{\rm{\;}} + {\rm{\;sin\theta }}}}{{{\rm{cos\theta }}}} = {\rm{\;}}\frac{{1{\rm{\;}} + {\rm{\;}}\frac{{21}}{{29}}}}{{\frac{{20}}{{29}}}} = {\rm{\;}}\frac{{50}}{{20}} = {\rm{\;}}\frac{5}{2})$

Discussion

153.	Find the value of 3(sin12A + 3sin10A + 3 sin8A + sin6A + 1) if (1/secA) + (1/sec2A) = 1?1). 32). 43). 84). 6
Answer» ⇒ (1/secA) + (1/sec²A) = 1 ⇒ COSA + cos²A = 1 ⇒ cosA = sin²A ⇒ 3((sin⁴A) ³ + (sin²) ³ + 3sin⁶A(sin⁴A + sin²A) + 1) ⇒ 3((sin⁴A + sin²A) ³ + 1) ⇒ 3((cos²A + cosA) ³ + 1) ∴ 3(1 + 1) = 6

153.

Find the value of 3(sin12A + 3sin10A + 3 sin8A + sin6A + 1) if (1/secA) + (1/sec2A) = 1?1). 32). 43). 84). 6

Answer»

⇒ (1/secA) + (1/sec²A) = 1

⇒ COSA + cos²A = 1

⇒ cosA = sin²A

⇒ 3((sin⁴A) ³ + (sin²) ³ + 3sin⁶A(sin⁴A + sin²A) + 1)

⇒ 3((sin⁴A + sin²A) ³ + 1)

⇒ 3((cos²A + cosA) ³ + 1)

∴ 3(1 + 1) = 6

Discussion

154.	ABC is a triangle with ∠BAC = 90°, then what is the value of sin2A + sin2B + sin2C?1). 12). 23). 34). 4
Answer» sin²A + sin²B + sin²C = sin²90° + sin²(90° – C) + sin²C (? In a RIGHT angled triangle SUM of other two angles = 90°) = 1 + cos²C + sin²C (? sin²θ + cos²θ = 1) = 1 + 1 = 2 ∴ sin²A + sin²B + sin²C = 2

154.

ABC is a triangle with ∠BAC = 90°, then what is the value of sin2A + sin2B + sin2C?1). 12). 23). 34). 4

Answer»

sin²A + sin²B + sin²C

= sin²90° + sin²(90° – C) + sin²C

(? In a RIGHT angled triangle SUM of other two angles = 90°)

= 1 + cos²C + sin²C

(? sin²θ + cos²θ = 1)

= 1 + 1 = 2

∴ sin²A + sin²B + sin²C = 2

Discussion

155.	1). x/22). 2x3). x4). 1/2x
Answer» By SOLVING the equation : $(\RIGHTARROW {\rm{SEC\theta }} = \FRAC{{4{{\rm{X}}^2} + 1}}{{4{\rm{x}}}})$ Using right angle theorem: ⇒ sec

155.

1). x/22). 2x3). x4). 1/2x

Answer»

By SOLVING the equation :

$(\RIGHTARROW {\rm{SEC\theta }} = \FRAC{{4{{\rm{X}}^2} + 1}}{{4{\rm{x}}}})$

Using right angle theorem:

⇒ sec

Discussion

156.	If x, y are acute angles 0 < x + y < 90° and sin(2x – 20°) = cos(2y + 20°), then the value of tan(x + y) is1). 1/√32). √3/23). √34). 1
Answer» We know that, cos θ = sin (90° - θ) GIVEN expression, sin (2X – 20°) = cos (2Y + 20°) ⇒ sin (2x – 20°) = sin [90° - (2y + 20°)] ⇒ (2x -20°) = n

156.

If x, y are acute angles 0 < x + y < 90° and sin(2x – 20°) = cos(2y + 20°), then the value of tan(x + y) is1). 1/√32). √3/23). √34). 1

Answer»

We know that, cos θ = sin (90° - θ)

GIVEN expression,

sin (2X – 20°) = cos (2Y + 20°)

⇒ sin (2x – 20°) = sin [90° - (2y + 20°)]

⇒ (2x -20°) = n

Discussion

157.	What is the simplified value of (sec4 A - tan2 A) - (tan4 A + sec2 A)?1). -12). -1/23). 04). 1
Answer» Put A = 0 ⇒ sec A = 1 and tan A = 0 Now put this in the required expression ⇒ (sec⁴ A – tan² A) - (tan⁴ A + sec² A) ⇒ (1) - (1) = 0 ∴ The SIMPLIFIED value of (sec⁴ A - tan² A) - (tan⁴ A + sec² A) is 0

157.

What is the simplified value of (sec4 A - tan2 A) - (tan4 A + sec2 A)?1). -12). -1/23). 04). 1

Answer»

Put A = 0

⇒ sec A = 1 and tan A = 0

Now put this in the required expression

⇒ (sec⁴ A – tan² A) - (tan⁴ A + sec² A)

⇒ (1) - (1) = 0

∴ The SIMPLIFIED value of (sec⁴ A - tan² A) - (tan⁴ A + sec² A) is 0

Discussion

158.	Solve: 1 + tan2 θ = ?1). cos2 θ2). sec2 θ3). tan2 θ4). 2
Answer»

Discussion

159.	1). 12). -13). 04). 2
Answer» $(\frac{{\left[ {4cos\left( {90 - A} \RIGHT).si{n^3}\left( {90\; + \;A} \right)} \right] - \left[ {4sin\left( {90\; + \;A} \right).CO{s^3}\left( {90 - A} \right)} \right]}}{{cos\left( {\frac{{180\; + \;8A}}{2}} \right)\;}})$ $(\Rightarrow \frac{{\left( {4sinA.co{s^3}A} \right) - \left( {4cosA.si{n^3}A} \right)}}{{ - sin4A}})$ $(\Rightarrow \frac{{\left( {4sinA.co{s^3}A} \right) - \left( {4cosA.si{n^3}A} \right)}}{{ - 2sin2Acos2A}})$ $(\Rightarrow \frac{{\left( {4sinA.co{s^3}A} \right) - \left( {4cosA.si{n^3}A} \right)}}{{ - 4sinA.cosA.COS2A}})$ $(\Rightarrow \frac{{co{s^2}A - si{n^2}A\;}}{{ - cos2A}})$ Since, cos2A = cos²A – sin²A $(\Rightarrow \frac{{co{s^2}A - si{n^2}A\;}}{{-\left( {co{s^2}A-si{n^2}A} \right)}})$ ⇒ -1

159.

1). 12). -13). 04). 2

Answer»

$(\frac{{\left[ {4cos\left( {90 - A} \RIGHT).si{n^3}\left( {90\; + \;A} \right)} \right] - \left[ {4sin\left( {90\; + \;A} \right).CO{s^3}\left( {90 - A} \right)} \right]}}{{cos\left( {\frac{{180\; + \;8A}}{2}} \right)\;}})$

$(\Rightarrow \frac{{\left( {4sinA.co{s^3}A} \right) - \left( {4cosA.si{n^3}A} \right)}}{{ - sin4A}})$

$(\Rightarrow \frac{{\left( {4sinA.co{s^3}A} \right) - \left( {4cosA.si{n^3}A} \right)}}{{ - 2sin2Acos2A}})$

$(\Rightarrow \frac{{\left( {4sinA.co{s^3}A} \right) - \left( {4cosA.si{n^3}A} \right)}}{{ - 4sinA.cosA.COS2A}})$

$(\Rightarrow \frac{{co{s^2}A - si{n^2}A\;}}{{ - cos2A}})$

Since, cos2A = cos²A – sin²A

$(\Rightarrow \frac{{co{s^2}A - si{n^2}A\;}}{{-\left( {co{s^2}A-si{n^2}A} \right)}})$

⇒ -1

Discussion

160.	If \(sin\left( {A - B} \right) = \frac{1}{2}\) and \(\cos \left( {A + B} \right) = \frac{1}{2}\) where A > B > 0 and A + B is an acute angle, then the value of angle B is1). π/62). π/123). π/44). π/2
Answer» GIVEN, $(sin\left( {A - B} \right) = \frac{1}{2})$ $(\because\sin \frac{\pi }{6} = \frac{1}{2})$ ∴ A – B = π/6…..(1) And, $(cos\left( {A + B} \right) = \frac{1}{2})$ $(\because\cos \frac{\pi }{3} = \frac{1}{2})$ ∴ A + B = π/3…..(2) (2) – (1) A + B – A + B = π/3 – π/6 ⇒ 2B = π/6 ⇒ B = π/12

160.

If $sin\left( {A - B} \right) = \frac{1}{2}$ and $\cos \left( {A + B} \right) = \frac{1}{2}$ where A > B > 0 and A + B is an acute angle, then the value of angle B is1). π/62). π/123). π/44). π/2

Answer»

GIVEN,

$(sin\left( {A - B} \right) = \frac{1}{2})$

$(\because\sin \frac{\pi }{6} = \frac{1}{2})$

∴ A – B = π/6…..(1)

And,

$(cos\left( {A + B} \right) = \frac{1}{2})$

$(\because\cos \frac{\pi }{3} = \frac{1}{2})$

∴ A + B = π/3…..(2)

(2) – (1)

A + B – A + B = π/3 – π/6

⇒ 2B = π/6

⇒ B = π/12

Discussion

161.	What is the minimum value of 4 sin2 θ + 6 cos2 θ?1). 42). 53). 34). 1
Answer» Formula: sin² θ + cos² θ = 1 ⇒ Let the minimum value be x. ⇒ x = 4 sin² θ + 6 cos² θ ⇒ x = 4 (1 - cos² θ) + 6 cos² θ ⇒ x = 4 - 4 cos² θ + 6 cos² θ ⇒ x = 4 + 2 cos² θ ⇒ Value of cos θ is minimum (0) at 90 ⇒ x = 4 + 2 × 0 ∴ x = 4

161.

What is the minimum value of 4 sin2 θ + 6 cos2 θ?1). 42). 53). 34). 1

Answer»

Formula: sin² θ + cos² θ = 1

⇒ Let the minimum value be x.

⇒ x = 4 sin² θ + 6 cos² θ

⇒ x = 4 (1 - cos² θ) + 6 cos² θ

⇒ x = 4 - 4 cos² θ + 6 cos² θ

⇒ x = 4 + 2 cos² θ

⇒ Value of cos θ is minimum (0) at 90

⇒ x = 4 + 2 × 0

∴ x = 4

Discussion

162.	The angle of elevation of the top of a tower from point P is 15°. Point P is on the ground level. Point P is at a distance of 100 meters from the foot of the tower. Find the height of the tower.1). 26.8 meters2). 25.9 meters3). 30.5 meters4). 29.9 meters
Answer»

Discussion

163.	Value of (tan1° tan2° tan3° ………..tan89°) is:1). 12). 893). Undefined4). 0
Answer» We KNOW that, TAN(90 - θ) = cot(θ) tan 89 can be written as cot 1 [tan(90 - 1) = cot 1] So, tan 88 = cot 2 tan 87 = cot 3 up to tan 46 = cot 44 then middle one is tan 45 = 1 Hence the expression becomes tan 1 × tan 2...........tan 44 × tan 45 × cot 44..............cot 1 as tanθ × cotθ = 1 hence tan and cot GETS cancelled Thus the value that remain is tan 45 i.e. 1

163.

Value of (tan1° tan2° tan3° ………..tan89°) is:1). 12). 893). Undefined4). 0

Answer»

We KNOW that, TAN(90 - θ) = cot(θ)

tan 89 can be written as cot 1 [tan(90 - 1) = cot 1]

So,

tan 88 = cot 2

tan 87 = cot 3

up to tan 46 = cot 44

then middle one is tan 45 = 1

Hence the expression becomes

tan 1 × tan 2...........tan 44 × tan 45 × cot 44..............cot 1

as tanθ × cotθ = 1 hence tan and cot GETS cancelled

Thus the value that remain is tan 45 i.e. 1

Discussion

164.	If √(1 + cot2A) = x, then the value of x is?1). secA2). sinA3). cosecA4). cosA
Answer» ⇒ x = √(1 + cot²A) ⇒ x = √(1 + (cos²A/sin²A))[? cotA = cosA/sinA] ⇒ x = √((sin²A + cos²A)/sin²A) ⇒ x = √(1/sin²A)[? sin²A + cos²A = 1] ⇒ x = 1/sinA = COSECA

164.

If √(1 + cot2A) = x, then the value of x is?1). secA2). sinA3). cosecA4). cosA

Answer»

⇒ x = √(1 + cot²A)

⇒ x = √(1 + (cos²A/sin²A))[? cotA = cosA/sinA]

⇒ x = √((sin²A + cos²A)/sin²A)

⇒ x = √(1/sin²A)[? sin²A + cos²A = 1]

⇒ x = 1/sinA = COSECA

Discussion

165.	If (cotA – tan A)/2 = x, then the value of x is1). tan2A2). cot2A3). tanA4). cotA
Answer» From the GIVEN data, COTA = cosA/SINA and tanA = sinA/cosA ⇒ [(cosA/sinA) - (sinA/cosA)]/2 = x ⇒ (cos²A - sin²A)/ (2sinAcosA) = x We know that, COS2A = cos²A - sin²A and sin2A = 2 × sinA × cosA ⇒ cos2A/sin2A = x ⇒ x = cot2A

165.

If (cotA – tan A)/2 = x, then the value of x is1). tan2A2). cot2A3). tanA4). cotA

Answer»

From the GIVEN data,

COTA = cosA/SINA and tanA = sinA/cosA

⇒ [(cosA/sinA) - (sinA/cosA)]/2 = x

⇒ (cos²A - sin²A)/ (2sinAcosA) = x

We know that, COS2A = cos²A - sin²A and sin2A = 2 × sinA × cosA

⇒ cos2A/sin2A = x

⇒ x = cot2A

Discussion

166.	The value of \(\left( {{{\sin }^2}16\frac{{1^\circ }}{2} + {{\sin }^2}73\frac{{1^\circ }}{2}} \right)\) is1). 12). 23). 04). 4
Answer» $(\Rightarrow \;\LEFT( {{{\SIN }^2}16\FRAC{{1^\circ }}{2} + {{\sin }^2}73\frac{{1^\circ }}{2}} \right)\; = \;\left( {{{\sin }^2}16\frac{{1^\circ }}{2} + {{\sin }^2}(90 - 16\;\frac{{1^\circ }}{2}} \right)\; = \;\left( {{{\sin }^2}16\frac{{1^\circ }}{2} + {{\cos }^2}16\frac{{1^\circ }}{2}} \right))$ (? sin²x + cos²x = 1) = 1

166.

The value of $\left( {{{\sin }^2}16\frac{{1^\circ }}{2} + {{\sin }^2}73\frac{{1^\circ }}{2}} \right)$ is1). 12). 23). 04). 4

Answer»

$(\Rightarrow \;\LEFT( {{{\SIN }^2}16\FRAC{{1^\circ }}{2} + {{\sin }^2}73\frac{{1^\circ }}{2}} \right)\; = \;\left( {{{\sin }^2}16\frac{{1^\circ }}{2} + {{\sin }^2}(90 - 16\;\frac{{1^\circ }}{2}} \right)\; = \;\left( {{{\sin }^2}16\frac{{1^\circ }}{2} + {{\cos }^2}16\frac{{1^\circ }}{2}} \right))$

(? sin²x + cos²x = 1)

= 1

Discussion

167.	What is the value of \(\frac{{\cos 2{\rm{A\;}} + {\rm{\;}}2{{\cos }^2}{\rm{A\;}} - {\rm{\;}}2\cos 2{\rm{A}}\cos {\rm{A}}}}{{\sin 2{\rm{A\;}} - {\rm{\;}}2{{\sin }^2}{\rm{A}}\sin 2{\rm{A}}}}?\)1). (8 - 2√3) / √32). (2 + √3) / √33). (2√3 - 2) / √34). (2 - 2√3) / √3
Answer» $(\frac{{\COS 2{\rm{A\;}} + {\rm{\;}}2{{\cos }^2}{\rm{A\;}} - {\rm{\;}}2\cos 2{\rm{A}}\cos {\rm{A}}}}{{\sin 2{\rm{A\;}} - {\rm{\;}}2{{\sin }^2}{\rm{A}}\sin 2{\rm{A}}}})$ PUT A = 30° $(\Rightarrow \frac{{\cos 60{\rm{\;}} + {\rm{\;}}2{{\cos }^2}30 - {\rm{\;}}2\cos 60\cos 30}}{{\sin 60{\rm{\;}} - {\rm{\;}}2{{\sin }^2}30\sin 60}})$ $(\Rightarrow \frac{{\frac{1}{2}{\rm{\;}} + {\rm{\;}}2{\rm{\;}} \times {\rm{\;}}\frac{3}{4}\; - {\rm{\;}}2{\rm{\;}} \times {\rm{\;}}\frac{{\sqrt 3 }}{2}\; \times {\rm{\;}}\frac{1}{2}{\rm{\;}}}}{{\frac{{\sqrt 3 }}{2}{\rm{\;}} - {\rm{\;}}2 \times {\rm{\;}}\frac{1}{4}\; \times {\rm{\;}}\frac{{\sqrt 3 }}{2}}})$ ⇒ (2 - √3/2) / (√3/2 - √3/4) = (8 - 2√3) /√3

167.

What is the value of $\frac{{\cos 2{\rm{A\;}} + {\rm{\;}}2{{\cos }^2}{\rm{A\;}} - {\rm{\;}}2\cos 2{\rm{A}}\cos {\rm{A}}}}{{\sin 2{\rm{A\;}} - {\rm{\;}}2{{\sin }^2}{\rm{A}}\sin 2{\rm{A}}}}?$1). (8 - 2√3) / √32). (2 + √3) / √33). (2√3 - 2) / √34). (2 - 2√3) / √3

Answer»

$(\frac{{\COS 2{\rm{A\;}} + {\rm{\;}}2{{\cos }^2}{\rm{A\;}} - {\rm{\;}}2\cos 2{\rm{A}}\cos {\rm{A}}}}{{\sin 2{\rm{A\;}} - {\rm{\;}}2{{\sin }^2}{\rm{A}}\sin 2{\rm{A}}}})$

PUT A = 30°

$(\Rightarrow \frac{{\cos 60{\rm{\;}} + {\rm{\;}}2{{\cos }^2}30 - {\rm{\;}}2\cos 60\cos 30}}{{\sin 60{\rm{\;}} - {\rm{\;}}2{{\sin }^2}30\sin 60}})$

$(\Rightarrow \frac{{\frac{1}{2}{\rm{\;}} + {\rm{\;}}2{\rm{\;}} \times {\rm{\;}}\frac{3}{4}\; - {\rm{\;}}2{\rm{\;}} \times {\rm{\;}}\frac{{\sqrt 3 }}{2}\; \times {\rm{\;}}\frac{1}{2}{\rm{\;}}}}{{\frac{{\sqrt 3 }}{2}{\rm{\;}} - {\rm{\;}}2 \times {\rm{\;}}\frac{1}{4}\; \times {\rm{\;}}\frac{{\sqrt 3 }}{2}}})$

⇒ (2 - √3/2) / (√3/2 - √3/4) = (8 - 2√3) /√3

Discussion

168.	1). 02). -13). 14). 2
Answer» secθ $( = \sqrt {2 + \sqrt {2 + \sqrt {2 +\ldots\ldots\ldots\ldots \INFTY } } } )$ ∴ secθ = $(\sqrt {2 + {\RM{sec\theta }}} )$ sec²θ = 2 + secθ sec²θ - secθ - 2 = 0 secθ $(= \frac{{1 \pm \sqrt {1 + 8} }}{2})$ secθ $( = \frac{{1 \pm 3}}{2})$ ∴ secθ = 2, -1 [secθ = -1 not possible] ∴ θ = 60° for 0 < θ <$(\frac{{\rm{\PI }}}{2})$ cosθ (1 +2cosθ) = cos60°(1 +2×cos60°) ∴ $( = \frac{1}{2}\left\{ {1 + 2 \times \frac{1}{2}} \RIGHT\} = 1)$ Shortcut: - Secθ = 2 Cosθ $( = \frac{1}{2})$ ∴ θ = 60° cos60°(1 + 2 cos60°) $( = \frac{1}{2}\left\{ {1 + 2 \times \frac{1}{2}} \right\})$ = 1

168.

1). 02). -13). 14). 2

Answer»

secθ $( = \sqrt {2 + \sqrt {2 + \sqrt {2 +\ldots\ldots\ldots\ldots \INFTY } } } )$

∴ secθ = $(\sqrt {2 + {\RM{sec\theta }}} )$

sec²θ = 2 + secθ

sec²θ - secθ - 2 = 0

secθ $(= \frac{{1 \pm \sqrt {1 + 8} }}{2})$

secθ $( = \frac{{1 \pm 3}}{2})$

∴ secθ = 2, -1

[secθ = -1 not possible]

∴ θ = 60° for 0 < θ <$(\frac{{\rm{\PI }}}{2})$

cosθ (1 +2cosθ) = cos60°(1 +2×cos60°)

∴ $( = \frac{1}{2}\left\{ {1 + 2 \times \frac{1}{2}} \RIGHT\} = 1)$

Shortcut: -

Secθ = 2

Cosθ $( = \frac{1}{2})$

∴ θ = 60°

cos60°(1 + 2 cos60°)

$( = \frac{1}{2}\left\{ {1 + 2 \times \frac{1}{2}} \right\})$

= 1

Discussion

169.	If a cot θ= b, the value of \(\frac{{a\sin \theta \; + \;b\cos \theta }}{{a\sin \theta - b\cos \theta }}\)1). \(\frac{{a - b}}{{{a^2}\; + \;{b^2}}}\)2). \(\frac{{a\; + \;b}}{{{a^2}\; + \;{b^2}}}\)3). \(\frac{{{a^2}\; + \;{b^2}}}{{{a^2} - {b^2}}}\)4). \(\frac{{{a^2} - {b^2}}}{{{a^2}\; + \;{b^2}}}\)
Answer» a COT θ= b tanθ = a/b $(\frac{{a\SIN \theta \; + \;b\cos \theta }}{{a\sin \theta - b\cos \theta }} = \;\frac{{a(\sin \theta /cos\theta ) - b(\cos \theta /cos\theta )}}{{a(\sin \theta /cos\theta )\; + \;b(\cos \theta /cos\theta )}}\; = \frac{{a\tan \theta \; + \;b}}{{a\tan \theta - b}} = \frac{{a\; \times \;\LEFT( {\frac{a}{b}} \right)\; + \;b}}{{a\; \times \;\left( {\frac{a}{b}} \right) - b}} = \;\frac{{{a^2}\; + \;{b^2}}}{{{a^2} - {b^2}}})$

169.

If a cot θ= b, the value of $\frac{{a\sin \theta \; + \;b\cos \theta }}{{a\sin \theta - b\cos \theta }}$1). $\frac{{a - b}}{{{a^2}\; + \;{b^2}}}$2). $\frac{{a\; + \;b}}{{{a^2}\; + \;{b^2}}}$3). $\frac{{{a^2}\; + \;{b^2}}}{{{a^2} - {b^2}}}$4). $\frac{{{a^2} - {b^2}}}{{{a^2}\; + \;{b^2}}}$

Answer»

a COT θ= b

tanθ = a/b

$(\frac{{a\SIN \theta \; + \;b\cos \theta }}{{a\sin \theta - b\cos \theta }} = \;\frac{{a(\sin \theta /cos\theta ) - b(\cos \theta /cos\theta )}}{{a(\sin \theta /cos\theta )\; + \;b(\cos \theta /cos\theta )}}\; = \frac{{a\tan \theta \; + \;b}}{{a\tan \theta - b}} = \frac{{a\; \times \;\LEFT( {\frac{a}{b}} \right)\; + \;b}}{{a\; \times \;\left( {\frac{a}{b}} \right) - b}} = \;\frac{{{a^2}\; + \;{b^2}}}{{{a^2} - {b^2}}})$

Discussion

170.	The value of 1/(1 – cot2 α) + 1/(1 – tan2 α) is equal to1). -1/22). -13). 14). ½
Answer» cotα = cosα/sinα, tanα = sinα/cosα 1/(1 – cot² α) + 1/(1 – tan² α) = [sin² α/(sin² α – cos² α)] + [cos² α/( cos² α - sin² α)] = [sin² α/(sin² α – cos² α)] - [cos² α/(sin² α – cos² α)] = 1

170.

The value of 1/(1 – cot2 α) + 1/(1 – tan2 α) is equal to1). -1/22). -13). 14). ½

Answer»

cotα = cosα/sinα, tanα = sinα/cosα

1/(1 – cot² α) + 1/(1 – tan² α) = [sin² α/(sin² α – cos² α)] + [cos² α/( cos² α - sin² α)] =

[sin² α/(sin² α – cos² α)] - [cos² α/(sin² α – cos² α)] = 1

Discussion

171.	What is the value of \(\frac{{\cot x}}{{1 - \tan x}} + \frac{{\tan x}}{{1 - \cot x}}?\)1). sin x cos x + 12). sec x cosec x + 13). tan x cot x + 14). sec2x cosec2 x + 1
Answer» $({{\COT x}}{{1 - \tan x}} + \frac{{\tan x}}{{1 - \cot x}} = \frac{{cosx}}{{\frac{{sinx}}{{1 - \frac{{sinx}}{{cosx}}}}}} + \frac{{sinx}}{{\frac{{cosx}}{{1 - \frac{{cosx}}{{sinx}}}}}} = \frac{{cosx}}{{\frac{{sinx}}{{\frac{{cosx - sinx}}{{cosx}}}}}} + \frac{{sinx}}{{\frac{{cosx}}{{\frac{{sinx - cosx}}{{sinx}}}}}})$ $(= \frac{{co{s^2}x}}{{sinx\left( {cosx - sinx} \RIGHT)}} + \frac{{si{n^2}x}}{{cosx\left( {sinx - cosx} \right)}})$ $(= \frac{{co{s^2}x}}{{sinx\left( {cosx - sinx} \right)}} - \frac{{si{n^2}x}}{{cosx\left( {cosx - sinx} \right)}})$ $(= \frac{{co{s^3}x - si{n^3}x}}{{\left( {cosx - sinx} \right)sinx\;cosx}} = \frac{{\left[ {\left( {cosx - sinx} \right)\left( {si{n^2}x +co{s^2}x\; + \;sinx\;cosx} \right)} \right]}}{{\left( {cosx - sinx} \right)\sin cosx}}\;)$ $(= \frac{{1 +sinx\;cosx}}{{SINXCOSX}})$ $(= COSECX\;secx + 1)$

171.

What is the value of $\frac{{\cot x}}{{1 - \tan x}} + \frac{{\tan x}}{{1 - \cot x}}?$1). sin x cos x + 12). sec x cosec x + 13). tan x cot x + 14). sec2x cosec2 x + 1

Answer»

$({{\COT x}}{{1 - \tan x}} + \frac{{\tan x}}{{1 - \cot x}} = \frac{{cosx}}{{\frac{{sinx}}{{1 - \frac{{sinx}}{{cosx}}}}}} + \frac{{sinx}}{{\frac{{cosx}}{{1 - \frac{{cosx}}{{sinx}}}}}} = \frac{{cosx}}{{\frac{{sinx}}{{\frac{{cosx - sinx}}{{cosx}}}}}} + \frac{{sinx}}{{\frac{{cosx}}{{\frac{{sinx - cosx}}{{sinx}}}}}})$

$(= \frac{{co{s^2}x}}{{sinx\left( {cosx - sinx} \RIGHT)}} + \frac{{si{n^2}x}}{{cosx\left( {sinx - cosx} \right)}})$

$(= \frac{{co{s^2}x}}{{sinx\left( {cosx - sinx} \right)}} - \frac{{si{n^2}x}}{{cosx\left( {cosx - sinx} \right)}})$

$(= \frac{{co{s^3}x - si{n^3}x}}{{\left( {cosx - sinx} \right)sinx\;cosx}} = \frac{{\left[ {\left( {cosx - sinx} \right)\left( {si{n^2}x +co{s^2}x\; + \;sinx\;cosx} \right)} \right]}}{{\left( {cosx - sinx} \right)\sin cosx}}\;)$

$(= \frac{{1 +sinx\;cosx}}{{SINXCOSX}})$

$(= COSECX\;secx + 1)$

Discussion

172.	ΔDEF is right angled at E. If ∠F = 45°, then what is the value of 2 Sin F x Cot F?1). √22). 23). 1/√24). 1/2
Answer» GIVEN, ∠F = 45° 2 sin F X cot F ⇒ 2 sin F × (COS F/sin F) ⇒ 2 cos F ⇒ 2 cos 45° ⇒ 2 × (1/√2)(? cos 45° = 1/√2) ⇒ √2

172.

ΔDEF is right angled at E. If ∠F = 45°, then what is the value of 2 Sin F x Cot F?1). √22). 23). 1/√24). 1/2

Answer»

GIVEN, ∠F = 45°

2 sin F X cot F

⇒ 2 sin F × (COS F/sin F)

⇒ 2 cos F

⇒ 2 cos 45°

⇒ 2 × (1/√2)(? cos 45° = 1/√2)

⇒ √2

Discussion

173.	1). (1 + sinA)/(1 - sinA)2). √[(1 - sinA)/(1 + sinA)]3). (1 - sinA)/(1 + sinA)4). √[(1 + sinA)/(1 - sinA)]
Answer» (secA - TANA)² = (1/cos A – sinA/cosA)² = [(1 – sinA)/cosA]² = (1 – sinA)²/cos²A = (1 – sinA)²/[(1 – sinA)(1 + sinA)] = (1 – sinA)/(1 + sinA) ∴ x = (1 – sinA)/(1 + sinA)

173.

1). (1 + sinA)/(1 - sinA)2). √[(1 - sinA)/(1 + sinA)]3). (1 - sinA)/(1 + sinA)4). √[(1 + sinA)/(1 - sinA)]

Answer»

(secA - TANA)²

= (1/cos A – sinA/cosA)²

= [(1 – sinA)/cosA]²

= (1 – sinA)²/cos²A

= (1 – sinA)²/[(1 – sinA)(1 + sinA)]

= (1 – sinA)/(1 + sinA)

∴ x = (1 – sinA)/(1 + sinA)

Discussion

174.	What is the simplified value of sin2(90 – θ) – [{sin(90 – θ)sin θ}/tanθ]?1). 12). cosec θ3). 04). cos θ
Answer» We KNOW that: SIN(90 – θ) = cosθ So, sin²(90 – θ) – [{sin(90 – θ)sinθ}/tanθ] = cos²θ – [(cosθ sinθ)/(sinθ/cosθ)] ⇒ cos²θ – cos²θ ∴ 0

174.

What is the simplified value of sin2(90 – θ) – [{sin(90 – θ)sin θ}/tanθ]?1). 12). cosec θ3). 04). cos θ

Answer»

We KNOW that: SIN(90 – θ) = cosθ

So, sin²(90 – θ) – [{sin(90 – θ)sinθ}/tanθ] = cos²θ – [(cosθ sinθ)/(sinθ/cosθ)]

⇒ cos²θ – cos²θ

∴ 0

Discussion

175.	1). 452). 2003). 1004). 136.6
Answer»

Discussion

176.	If sin(3θ)sec(2θ) = 1, then what is the value of [3tan2(5θ/2) – 1]?1). 02). 13). 24). 3
Answer» ⇒ sin3θ = 1/sec2θ ⇒ cos(π/2 – 3θ) = cos2θ ⇒ π/2 – 3θ = 2θ ⇒ π/2 = 5θ ⇒ θ = π/10 To FIND: [3tan²(5θ/2) – 1] ⇒ 3tan²(π/4) – 1 ⇒ 3 × 1 – 1 = 2 ∴ the CORRECT option is 3)

176.

If sin(3θ)sec(2θ) = 1, then what is the value of [3tan2(5θ/2) – 1]?1). 02). 13). 24). 3

Answer»

⇒ sin3θ = 1/sec2θ

⇒ cos(π/2 – 3θ) = cos2θ

⇒ π/2 – 3θ = 2θ

⇒ π/2 = 5θ

⇒ θ = π/10

To FIND: [3tan²(5θ/2) – 1]

⇒ 3tan²(π/4) – 1

⇒ 3 × 1 – 1 = 2

∴ the CORRECT option is 3)

Discussion

177.	1). 182). 123). 244). 23
Answer»

Discussion

178.	If (1 + sec A)/tan A = x, then x is1). cot (A/2)2). tan(A/2)3). cosec(A/2)4). sec(A/2)
Answer» ⇒ (1 + SEC A)/tanA $(= \frac{{1{\rm{}} + {\rm{}}\frac{1}{{{\rm{cosA}}}}}}{{{\rm{tanA}}}} = \frac{{1{\rm{}} + {\rm{cosA}}}}{{{\rm{sinA}}}} = \frac{{1{\rm{}} + {\rm{cosA}}}}{{\SQRT {1 - {\rm{co}}{{\rm{s}}^2}{\rm{A}}} }})$ $(\begin{ARRAY}{l} = \frac{{1{\rm{}} + {\rm{cosA}}}}{{\sqrt {\LEFT( {1{\rm{}} + {\rm{cosA}}} \right)\left( {1 - {\rm{cosA}}} \right)} }}\\ = \frac{{\sqrt {1{\rm{}} + {\rm{cosA}}} }}{{\sqrt {1 - {\rm{cosA}}} }} \end{array})$ = cot(A/2)

178.

If (1 + sec A)/tan A = x, then x is1). cot (A/2)2). tan(A/2)3). cosec(A/2)4). sec(A/2)

Answer»

⇒ (1 + SEC A)/tanA $(= \frac{{1{\rm{}} + {\rm{}}\frac{1}{{{\rm{cosA}}}}}}{{{\rm{tanA}}}} = \frac{{1{\rm{}} + {\rm{cosA}}}}{{{\rm{sinA}}}} = \frac{{1{\rm{}} + {\rm{cosA}}}}{{\SQRT {1 - {\rm{co}}{{\rm{s}}^2}{\rm{A}}} }})$

$(\begin{ARRAY}{l} = \frac{{1{\rm{}} + {\rm{cosA}}}}{{\sqrt {\LEFT( {1{\rm{}} + {\rm{cosA}}} \right)\left( {1 - {\rm{cosA}}} \right)} }}\\ = \frac{{\sqrt {1{\rm{}} + {\rm{cosA}}} }}{{\sqrt {1 - {\rm{cosA}}} }} \end{array})$

= cot(A/2)

Discussion

179.	If 5 secθ – 3 tanθ = 5, then what is the value of 5 tanθ – 3 secθ?1). 12). 23). 34). 4
Answer» 5 sec Θ - 3tan Θ = 5 ⇒ 5 √(tan² Θ + 1) = 5 + 3tanΘ ⇒ 25(tan² Θ + 1) = 25 + 9tan² Θ + 30 tanΘ ⇒ 25 tan² Θ – 9tan² Θ = 30 tanΘ ⇒ 16 tan² Θ - 30 tanΘ = 0 ⇒ 2tanΘ (8 tanΘ – 15) = 0 ⇒ tanΘ = 0 or tanΘ = 15/8 If TAN Θ = 0 5 sec Θ - 3 tan Θ = 5 ⇒ 5SEC Θ – 3 × 0 = 5 ⇒ sec Θ = 1 If tan Θ = 15/8 5sec Θ – 3 × 15/8 = 5 ⇒ 5 sec Θ = 5 + 45/8 ⇒ 5 secΘ = 85 /8 ⇒ sec Θ = 17/8 So, 5 tanΘ – 3 sec Θ = 5 × 0 – 3 × 1 = -3 Or, 5 × 15/8 – 3 × 17/8 = 75/8 - 51/8 = 24/8 = 3

179.

If 5 secθ – 3 tanθ = 5, then what is the value of 5 tanθ – 3 secθ?1). 12). 23). 34). 4

Answer»

5 sec Θ - 3tan Θ = 5

⇒ 5 √(tan² Θ + 1) = 5 + 3tanΘ

⇒ 25(tan² Θ + 1) = 25 + 9tan² Θ + 30 tanΘ

⇒ 25 tan² Θ – 9tan² Θ = 30 tanΘ

⇒ 16 tan² Θ - 30 tanΘ = 0

⇒ 2tanΘ (8 tanΘ – 15) = 0

⇒ tanΘ = 0 or tanΘ = 15/8

If TAN Θ = 0

5 sec Θ - 3 tan Θ = 5

⇒ 5SEC Θ – 3 × 0 = 5

⇒ sec Θ = 1

If tan Θ = 15/8

5sec Θ – 3 × 15/8 = 5

⇒ 5 sec Θ = 5 + 45/8

⇒ 5 secΘ = 85 /8

⇒ sec Θ = 17/8

So, 5 tanΘ – 3 sec Θ = 5 × 0 – 3 × 1 = -3

Or, 5 × 15/8 – 3 × 17/8 = 75/8 - 51/8 = 24/8 = 3

Discussion

180.	If 4a = sec θ and 4/a = tanθ, then \(8\left( {{a^2} - \frac{1}{{{a^2}}}} \right)\) is1). 1/162). 1/83). 1/24). 1/4
Answer» ⇒ 4a = sec θ ⇒ a = sec θ/4----1 ⇒ 4/a = tanθ ⇒ 1/a = tanθ/ 4----2 ⇒ We NEED to find the value of $(8\left( {{a^2} - \FRAC{1}{{{a^2}}}} \right))$ ⇒ $(8\left( {{a^2} - \frac{1}{{{a^2}}}} \right))$= 8(a – (1/a)) (a + (1/a)) ⇒ Putting the value of a and 1/a in above we get ⇒ $(8\left( {{a^2} - \frac{1}{{{a^2}}}} \right))$= 8 ((sec θ/4) - (tanθ/4)) × ((sec θ/4) + (tanθ/4)) ⇒ $(8\left( {{a^2} - \frac{1}{{{a^2}}}} \right))$= 8/16 (sec²θ - tan²θ) $(\THEREFORE \;8\left( {{a^2} - \frac{1}{{{a^2}}}} \right))$= 1/2

180.

If 4a = sec θ and 4/a = tanθ, then $8\left( {{a^2} - \frac{1}{{{a^2}}}} \right)$ is1). 1/162). 1/83). 1/24). 1/4

Answer»

⇒ 4a = sec θ

⇒ a = sec θ/4----1

⇒ 4/a = tanθ

⇒ 1/a = tanθ/ 4----2

⇒ We NEED to find the value of $(8\left( {{a^2} - \FRAC{1}{{{a^2}}}} \right))$

⇒ $(8\left( {{a^2} - \frac{1}{{{a^2}}}} \right))$= 8(a – (1/a)) (a + (1/a))

⇒ Putting the value of a and 1/a in above we get

⇒ $(8\left( {{a^2} - \frac{1}{{{a^2}}}} \right))$= 8 ((sec θ/4) - (tanθ/4)) × ((sec θ/4) + (tanθ/4))

⇒ $(8\left( {{a^2} - \frac{1}{{{a^2}}}} \right))$= 8/16 (sec²θ - tan²θ)

$(\THEREFORE \;8\left( {{a^2} - \frac{1}{{{a^2}}}} \right))$= 1/2

Discussion

181.	If secθ(cosθ + sinθ) = √2, then what is the value of 2sinθ/(cosθ - sinθ)?1). 3√22). 3/√23). 1/√24). √2
Answer» Simplifying the GIVEN equation, ⇒ 1 + TAN θ = √2 ⇒ tan θ = √2 - 1----(1) Now, 2sinθ/(cosθ - sinθ) Dividing by sinθ in both the numerator and denominator. ⇒ 2/(cotθ - 1) Putting the VALUE of cotθ from equation 1, ⇒ 2/[1/(√2 - 1) - 1] ⇒ (2√2 - 2)/(1 - √2 + 1) ⇒ (2√2 - 2)/(2 - √2) = √2 ∴ 2sinθ/(cosθ - sinθ) = √2

181.

If secθ(cosθ + sinθ) = √2, then what is the value of 2sinθ/(cosθ - sinθ)?1). 3√22). 3/√23). 1/√24). √2

Answer»

Simplifying the GIVEN equation,

⇒ 1 + TAN θ = √2

⇒ tan θ = √2 - 1----(1)

Now, 2sinθ/(cosθ - sinθ)

Dividing by sinθ in both the numerator and denominator.

⇒ 2/(cotθ - 1)

Putting the VALUE of cotθ from equation 1,

⇒ 2/[1/(√2 - 1) - 1]

⇒ (2√2 - 2)/(1 - √2 + 1)

⇒ (2√2 - 2)/(2 - √2) = √2

∴ 2sinθ/(cosθ - sinθ) = √2

Discussion

182.	If π sin θ = 1, π cosθ = 1, then the value of\(\left\{ {\sqrt 3 \tan \left( {\frac{2}{3}\theta } \right) + 1} \right\}\) is1). 02). √33). 24). 1
Answer» ⇒ given:π sin θ = 1-----1 and π cosθ = 1-----2 ⇒ dividing both the above equation we get ⇒ π sin θ/π cosθ = 1/1 ⇒ TAN θ = 1 ⇒ θ = 45 ⇒ Let x = $(\left\{ {\sqrt 3 \tan \left( {\FRAC{2}{3}\THETA } \right) + 1} \right\})$ ⇒ putting the VALUE of θ in ⇒ x = √3 tan (2/3)x 45) + 1 ⇒ x = √3 tan (30) + 1) ⇒ x = √3 x (1/√3) + 1 ∴ x = 2

182.

If π sin θ = 1, π cosθ = 1, then the value of$\left\{ {\sqrt 3 \tan \left( {\frac{2}{3}\theta } \right) + 1} \right\}$ is1). 02). √33). 24). 1

Answer»

⇒ given:π sin θ = 1-----1 and π cosθ = 1-----2

⇒ dividing both the above equation we get

⇒ π sin θ/π cosθ = 1/1

⇒ TAN θ = 1

⇒ θ = 45

⇒ Let x = $(\left\{ {\sqrt 3 \tan \left( {\FRAC{2}{3}\THETA } \right) + 1} \right\})$

⇒ putting the VALUE of θ in

⇒ x = √3 tan (2/3)x 45) + 1

⇒ x = √3 tan (30) + 1)

⇒ x = √3 x (1/√3) + 1

∴ x = 2

Discussion

183.	1). √[(1 + cosA)/(1 - cosA)]2). cosecA - cotA3). √[(1 - cosA)/2]4). √[(1 + cosA)/2]
Answer» As per the given data, COT (A/2) = x COS (A/2)/SIN (A/2) = x we know that: cos²A = (1 + cos 2A)/2 sin²A = (1 – cos 2A)/2 $(\frac{{\surd \left( {\frac{{1\; + \;cosA}}{2}} \right)}}{{\surd \left( {\frac{{1 - cosA}}{2}} \right)}})$ = x x = $(\frac{{\surd \left( {1\; + \;\cos A} \right)}}{{\surd \left( {1 - \cos A} \right)}})$

183.

1). √[(1 + cosA)/(1 - cosA)]2). cosecA - cotA3). √[(1 - cosA)/2]4). √[(1 + cosA)/2]

Answer»

As per the given data,

COT (A/2) = x

COS (A/2)/SIN (A/2) = x

we know that:

cos²A = (1 + cos 2A)/2

sin²A = (1 – cos 2A)/2

$(\frac{{\surd \left( {\frac{{1\; + \;cosA}}{2}} \right)}}{{\surd \left( {\frac{{1 - cosA}}{2}} \right)}})$ = x

x = $(\frac{{\surd \left( {1\; + \;\cos A} \right)}}{{\surd \left( {1 - \cos A} \right)}})$

Discussion

184.	The value of sin2 15° + sin2 20°+ sin2 25° + ….. + sin2 75° ---1). tan2 15° + sin2 20° + tan2 25° + …. tan2 75°2). cos2 15° + cos2 20° + cos2 25° + …. cos2 75°3). cot2 15° + cot2 20° + cot2 25° + …. cot2 75°4). sec2 15° + sec2 20° + sec2 25° + …. sec2 75°
Answer» sin²15° + sin²20° + sin²25° + ..... sin²75° = [COS (90° - 15°)]² + [cos (90° - 20°)]² + ..... [cos (90° - 75°)]² [cos (90 - θ) = sinθ] = cos²75° + cos²70° + cos²65° + ...... cos²15° = cos²15° + cos²20° + cos²25° + ..... cos²75°

184.

The value of sin2 15° + sin2 20°+ sin2 25° + ….. + sin2 75° ---1). tan2 15° + sin2 20° + tan2 25° + …. tan2 75°2). cos2 15° + cos2 20° + cos2 25° + …. cos2 75°3). cot2 15° + cot2 20° + cot2 25° + …. cot2 75°4). sec2 15° + sec2 20° + sec2 25° + …. sec2 75°

Answer»

sin²15° + sin²20° + sin²25° + ..... sin²75°

= [COS (90° - 15°)]² + [cos (90° - 20°)]² + ..... [cos (90° - 75°)]² [cos (90 - θ) = sinθ]

= cos²75° + cos²70° + cos²65° + ...... cos²15°

= cos²15° + cos²20° + cos²25° + ..... cos²75°

Discussion

185.	$\frac{{\cot {\rm{A\;}} + {\rm{\;}}\tan {\rm{B}}}}{{\cot {\rm{B\;}} + {\rm{\;}}\tan {\rm{A}}}}$ is equal to1). tanA cotB2). cotA tanB3). 14). None of these
Answer» $( \Rightarrow \frac{{{\rm{COTA}}+{\rm{COTB}}}}{{{\rm{cotB}} + {\rm{tanA}}}} = \frac{{\frac{{{\rm{COSA}}}}{{{\rm{SINA}}}}+\frac{{{\rm{sinB}}}}{{{\rm{cosB}}}}}}{{\frac{{{\rm{cosB}}}}{{{\rm{sinB}}}} + \frac{{{\rm{sinA\;}}}}{{{\rm{cosA}}}}}}=\frac{{\frac{{{\rm{cosAcosB}}+{\rm{sinAsinB}}}}{{{\rm{sinAcosB}}}}}}{{\frac{{{\rm{cosAcosB}}+{\rm{sinAsinB}}}}{{{\rm{sinBcosA}}}}}}=\frac{{{\rm{sinBcosA}}}}{{{\rm{sinAcosB}}}}={\rm{cotAtanB}})$

Discussion

186.	1). √3/22). 1/√33). 2/√34). √(2/3)
Answer» (5cosec²θ + 3cot²θ) = 21---- (1) We KNOW that: (cosec²θ – cot²θ) = 1 ∴ (3cosec²θ – 3cot²θ) = 3---- (2) ADDING equations (1) and (2) ⇒ 8cosec²θ = 24 ⇒ cosecθ = √3 ∴ cosθ = √(2/3)

186.

1). √3/22). 1/√33). 2/√34). √(2/3)

Answer»

(5cosec²θ + 3cot²θ) = 21---- (1)

We KNOW that:

(cosec²θ – cot²θ) = 1

∴ (3cosec²θ – 3cot²θ) = 3---- (2)

ADDING equations (1) and (2)

⇒ 8cosec²θ = 24

⇒ cosecθ = √3

∴ cosθ = √(2/3)

Discussion

187.	If cos Ɵ = √3/2 and 0° < Ɵ < 90°, then the value of cosec (Ɵ – 15°) is1). 12). √2 + √63). 1/√34). √2
Answer» If COS ? = √3/2 ⇒ ? = 30° ∴ COSEC (? – 15°) = cosec (30° – 15°) = cosec 15° = 1/(sin 15°) Since sin 15° = sin (45° – 30°) = sin 45° cos 30° – cos 45° sin 30° (? sin (A – B) = sin A cos B – cos A sin B) ∴ sin 15° $(= \frac{1}{{\sqrt 2 }} \times \frac{{\sqrt 3 }}{2} - \frac{1}{{\sqrt 2 }} \times \frac{1}{2} = \frac{{\sqrt 3 \;-1}}{{2\sqrt 2 }})$ ⇒ cosec 15° = $(\frac{{2\sqrt 2 }}{{\sqrt 3 - 1}} = \frac{{2\sqrt 2 }}{{\sqrt 3 - 1}} \times \frac{{\sqrt 3 + 1}}{{\sqrt 3 + 1}} = \;\frac{{2\sqrt 6 + 2\sqrt 2 }}{{3 - 1}} = \sqrt 2 + \sqrt 6 )$

187.

If cos Ɵ = √3/2 and 0° < Ɵ < 90°, then the value of cosec (Ɵ – 15°) is1). 12). √2 + √63). 1/√34). √2

Answer»

If COS ? = √3/2

⇒ ? = 30°

∴ COSEC (? – 15°) = cosec (30° – 15°) = cosec 15° = 1/(sin 15°)

Since sin 15° = sin (45° – 30°) = sin 45° cos 30° – cos 45° sin 30°

(? sin (A – B) = sin A cos B – cos A sin B)

∴ sin 15° $(= \frac{1}{{\sqrt 2 }} \times \frac{{\sqrt 3 }}{2} - \frac{1}{{\sqrt 2 }} \times \frac{1}{2} = \frac{{\sqrt 3 \;-1}}{{2\sqrt 2 }})$

⇒ cosec 15° = $(\frac{{2\sqrt 2 }}{{\sqrt 3 - 1}} = \frac{{2\sqrt 2 }}{{\sqrt 3 - 1}} \times \frac{{\sqrt 3 + 1}}{{\sqrt 3 + 1}} = \;\frac{{2\sqrt 6 + 2\sqrt 2 }}{{3 - 1}} = \sqrt 2 + \sqrt 6 )$

Discussion

188.	If $0^\circ \le \theta \le 90^\circ \;and\;sinx + \sqrt 3 \;cosx = 1$ then the value of x -1). 30°2). 45°3). 60°4). 90°
Answer»

Discussion

189.	If the value of θ = 30°, then value of tan2 θ + cot2θ1). 1/32). 4/33). 9/34). 10/3
Answer» The value of tan²θ + cot²θ = tan²30° + cot² 30° = (1/√3)² + (√3)² = 1/3 + 3 = (1 + 9)/3 = 10/3

189.

If the value of θ = 30°, then value of tan2 θ + cot2θ1). 1/32). 4/33). 9/34). 10/3

Answer»

The value of tan²θ + cot²θ = tan²30° + cot² 30°

= (1/√3)² + (√3)²

= 1/3 + 3 = (1 + 9)/3 = 10/3

Discussion

190.	If 5(sin A cos3 A + sin3 A cos A)3 – 2 sin3 A cos3 A = x sin3 2A, find the value of x.1). 1/32). 1/83). 2/34). 3/8
Answer» ? sin 2A = 2 sin A cos A ⇒ sin A cos A = (sin 2A)/2 Now, 5(sin A cos³ A + sin³ A cos A)³ – 2 sin³A cos³ A ⇒ 5[sin A cos A (cos² A + sin² A)]³ – 2 (sin A cos A)³ ⇒ 5(sin A cos A)³ – 2(sin A cos A)³[? (cos² A + sin² A) = 1] ⇒ 3(sin A cos A)³ ⇒ 3[(sin 2A)/2]³ ⇒ (3/8) sin³2A ∴ Comparing, we get, X = 3/8

190.

If 5(sin A cos3 A + sin3 A cos A)3 – 2 sin3 A cos3 A = x sin3 2A, find the value of x.1). 1/32). 1/83). 2/34). 3/8

Answer»

? sin 2A = 2 sin A cos A

⇒ sin A cos A = (sin 2A)/2

Now,

5(sin A cos³ A + sin³ A cos A)³ – 2 sin³A cos³ A

⇒ 5[sin A cos A (cos² A + sin² A)]³ – 2 (sin A cos A)³

⇒ 5(sin A cos A)³ – 2(sin A cos A)³[? (cos² A + sin² A) = 1]

⇒ 3(sin A cos A)³

⇒ 3[(sin 2A)/2]³

⇒ (3/8) sin³2A

∴ Comparing, we get, X = 3/8

Discussion

191.	1). 24/252). 7/253). 24/74). 25/7
Answer» Given SEC θ = 25/24, so Cos θ = 24/25 Sin θ can be given as ⇒ Sin θ = √(1 – COS2 θ) = √(1 – 242/252) = √{(625 – 57 6)/252} = √(72/252) = 7/25 ∴ Sin θ is 7/25

191.

1). 24/252). 7/253). 24/74). 25/7

Answer»

Given SEC θ = 25/24, so Cos θ = 24/25 Sin θ can be given as

⇒ Sin θ = √(1 – COS2 θ) = √(1 – 242/252) = √{(625 – 57 6)/252} = √(72/252) = 7/25

∴ Sin θ is 7/25

Discussion

192.	If cos2 θ - sin θ = 1/4, then what is the value of sin θ?1). -12). 1/23). 14). 3/2
Answer» Cos² θ = 1 - (SIN θ)² Let sin θ = X Equation becomes 1 - X² - X = 1/4 X² + X - 3/4 = 0 ∴ X = Sin θ = 1/2 Answer is (2)

192.

If cos2 θ - sin θ = 1/4, then what is the value of sin θ?1). -12). 1/23). 14). 3/2

Answer»

Cos² θ = 1 - (SIN θ)²

Let sin θ = X

Equation becomes

1 - X² - X = 1/4

X² + X - 3/4 = 0

∴ X = Sin θ = 1/2

Answer is (2)

Discussion

193.	If cos4 θ – sin4 θ = 3/5, then the value of 1 – 2 sin2 θ is1). 02). 3/53). 1/34). 4/3
Answer» cos⁴ θ – sin⁴ θ = (cos²θ)²– (sin² θ)² We know that, a² – b² = (a + b) (a – b) Hence, (cos²θ)²– (sin² θ)²= (cos² θ + sin² θ) (cos² θ - sin² θ) But, (cos² θ + sin² θ) = 1 Hence, (cos²θ)²– (sin² θ)²= (cos² θ - sin² θ) ALSO cos 2θ = cos² θ - sin² θ = 1 – 2sin² θ Therefore, cos⁴ θ – sin⁴ θ = 1 – 2sin² θ = 3/5

193.

If cos4 θ – sin4 θ = 3/5, then the value of 1 – 2 sin2 θ is1). 02). 3/53). 1/34). 4/3

Answer»

cos⁴ θ – sin⁴ θ = (cos²θ)²– (sin² θ)²

We know that,

a² – b² = (a + b) (a – b)

Hence,

(cos²θ)²– (sin² θ)²= (cos² θ + sin² θ) (cos² θ - sin² θ)

But, (cos² θ + sin² θ) = 1

Hence,

(cos²θ)²– (sin² θ)²= (cos² θ - sin² θ)

ALSO cos 2θ = cos² θ - sin² θ = 1 – 2sin² θ

Therefore,

cos⁴ θ – sin⁴ θ = 1 – 2sin² θ = 3/5

Discussion

194.	What is the value of (1/√3 + cot600)?1). 3√3/22). 7/2√33). (4 + √3)/2√34). 2/√3
Answer»

Discussion

195.	Find the square root of the expression: 16 sec4 a – 16 sec2 a1). 4 tan a sec2 a2). 16 tan a sec a3). 16 tan2 a sec a4). 4 tan a sec a
Answer» 16 sec⁴ a – 16 sec²a = 16(sec⁴ a – sec²a) ⇒ 16 sec² a (sec² a - 1) USING sec² a – TAN² a = 1 we get = 16(sec²a).(tan² a) TAKING square root of the expression we will get 4 tan a sec a.

195.

Find the square root of the expression: 16 sec4 a – 16 sec2 a1). 4 tan a sec2 a2). 16 tan a sec a3). 16 tan2 a sec a4). 4 tan a sec a

Answer»

16 sec⁴ a – 16 sec²a = 16(sec⁴ a – sec²a)

⇒ 16 sec² a (sec² a - 1)

USING sec² a – TAN² a = 1 we get

= 16(sec²a).(tan² a)

TAKING square root of the expression we will get 4 tan a sec a.

Discussion

196.	If sec (3x – 20)° = cosec (3y + 20)°, what is the value of tan (x + y)?1). 12). √33). 1/√34). 2√3
Answer» Since we know that SEC(x) = cosec (90 - x), From the given equation ⇒ sec (3x – 20) = sec{90 – (3y + 20)} comparing the terms in BRACKETS on LHS and RHS ⇒ 3x - 20 = 90 – (3y + 20) ⇒ x + y = 30 ∴ TAN(x + y) = tan(30) = 1/√3

196.

If sec (3x – 20)° = cosec (3y + 20)°, what is the value of tan (x + y)?1). 12). √33). 1/√34). 2√3

Answer»

Since we know that SEC(x) = cosec (90 - x), From the given equation

⇒ sec (3x – 20) = sec{90 – (3y + 20)} comparing the terms in BRACKETS on LHS and RHS

⇒ 3x - 20 = 90 – (3y + 20)

⇒ x + y = 30

∴ TAN(x + y) = tan(30) = 1/√3

Discussion

197.	What is the value of (1/√3 – sin 45°)?1). 1/2√32). (√2 – √3)/√63). 1 – √24). (4 – √3)/2
Answer» GIVEN that, (1/√3 – SIN 45°) ⇒ 1/(√3) – 1/(√2) ⇒ (√2 – √3)/(√6)

197.

What is the value of (1/√3 – sin 45°)?1). 1/2√32). (√2 – √3)/√63). 1 – √24). (4 – √3)/2

Answer»

GIVEN that,

(1/√3 – SIN 45°)

⇒ 1/(√3) – 1/(√2)

⇒ (√2 – √3)/(√6)

Discussion

198.	What is the value of (√2/3 - cosec60o)?1). (√6 - 6)/3√32). (2 - 2√3)/√33). (1 - √6)/√24). (4 - √3)/2√3
Answer» ∴ (√2/3 - cosec60^o) = √2/3 - 2/√3 = (√6- 6)/3√3

Discussion

199.	1). 12sin 30°2). 6sin 45°3). 12cos 30°4). 6sin 60°
Answer» TAN θ – cot θ = 0 ⇒ tan θ = cot θ ⇒ θ = 45^o √6(4SIN θ + 2COS θ) √6(4sin 45° + 2cos 45°) (? sin 45 = cos 45 = 1/√2) ⇒ √6(6/√2) = 6√3 = 12 × (√3/2) = 12 cos 30° ∴ √6(4sin θ + 2cos θ) = 12 cos 30°

199.

1). 12sin 30°2). 6sin 45°3). 12cos 30°4). 6sin 60°

Answer»

TAN θ – cot θ = 0

⇒ tan θ = cot θ

⇒ θ = 45^o

√6(4SIN θ + 2COS θ)

√6(4sin 45° + 2cos 45°)

(? sin 45 = cos 45 = 1/√2)

⇒ √6(6/√2) = 6√3 = 12 × (√3/2) = 12 cos 30°

∴ √6(4sin θ + 2cos θ) = 12 cos 30°

Discussion

200.	1). 702). 753). 804). 85
Answer»

Discussion

185.	\(\frac{{\cot {\rm{A\;}} + {\rm{\;}}\tan {\rm{B}}}}{{\cot {\rm{B\;}} + {\rm{\;}}\tan {\rm{A}}}}\) is equal to1). tanA cotB2). cotA tanB3). 14). None of these
Answer» $( \Rightarrow \frac{{{\rm{COTA}}+{\rm{COTB}}}}{{{\rm{cotB}} + {\rm{tanA}}}} = \frac{{\frac{{{\rm{COSA}}}}{{{\rm{SINA}}}}+\frac{{{\rm{sinB}}}}{{{\rm{cosB}}}}}}{{\frac{{{\rm{cosB}}}}{{{\rm{sinB}}}} + \frac{{{\rm{sinA\;}}}}{{{\rm{cosA}}}}}}=\frac{{\frac{{{\rm{cosAcosB}}+{\rm{sinAsinB}}}}{{{\rm{sinAcosB}}}}}}{{\frac{{{\rm{cosAcosB}}+{\rm{sinAsinB}}}}{{{\rm{sinBcosA}}}}}}=\frac{{{\rm{sinBcosA}}}}{{{\rm{sinAcosB}}}}={\rm{cotAtanB}})$

188.	If \(0^\circ \le \theta \le 90^\circ \;and\;sinx + \sqrt 3 \;cosx = 1\) then the value of x -1). 30°2). 45°3). 60°4). 90°
Answer»

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