232 + Interview Questions in QUANTITATIVE APTITUDE in TRIGONOMETRY Page 5 InterviewSolution

201.	ΔXYZ is right angled at Y. If sinX = 12/13, then what is the value of cosecZ ? 1). 5/122). 5/133). 13/54). 13/12
Answer» ⇒ sinX = 12/13 ⇒ YZ/XZ = 12/13 Let YZ = 12k and XZ = 13k By Pythagoras theorem, ⇒ XY = √(XZ² – YZ²) = √{(13k)² – (12k)²} = √(25k²) = 5k Now, ∴ COSEC Z = XZ/XY = 13k/5k = 13/5

201.

ΔXYZ is right angled at Y. If sinX = 12/13, then what is the value of cosecZ ? 1). 5/122). 5/133). 13/54). 13/12

Answer»

⇒ sinX = 12/13

⇒ YZ/XZ = 12/13

Let YZ = 12k and XZ = 13k

By Pythagoras theorem,

⇒ XY = √(XZ² – YZ²) = √{(13k)² – (12k)²} = √(25k²) = 5k

Now,

∴ COSEC Z = XZ/XY = 13k/5k = 13/5

Discussion

202.	For a man standing near a wall, the angles of elevation of the top of the wall from his head and foot is 30° and 60° respectively. If the height of the man is 200 cm, find the height of the wall.1). 2.4 m2). 2.7 m3). 3.0 m4). 3.5 m
Answer»

Discussion

203.	If sin θ + cos θ = 1, then sin θ cos θ is equal to:1). 02). 13). 24). -1
Answer» GIVEN, sin θ + cos θ = 1 squaring EQUATION, ⇒ sin²θ + 2sinθ. Cos θ + cos² θ = 1 ⇒ sin θ cos θ = 0

203.

If sin θ + cos θ = 1, then sin θ cos θ is equal to:1). 02). 13). 24). -1

Answer»

GIVEN,

sin θ + cos θ = 1

squaring EQUATION,

⇒ sin²θ + 2sinθ. Cos θ + cos² θ = 1

⇒ sin θ cos θ = 0

Discussion

204.	A tower standing on a horizontal plane subtends a certain angle at a point 80 m apart from the foot of the tower. On advancing 50 m towards it, the tower is found to subtend an angle twice as before. Find the height of the tower?1). 30 m2). 40 m3). 60 m4). 50 m
Answer» 3 ) 60

Discussion

205.	If sin A = x, then the value of tan A sec A1). \(\frac{x}{{\sqrt {1 + {x^2}} }}\)2). \(\sqrt {1 - {x^2}} \)3). \(\sqrt {\frac{{1 - {x^2}}}{{{x^2}}}}\)4). \(\frac{x}{{1 - {x^2}}}\)
Answer» sin A = Perpendicular/Hypotenuse = X/1 From PYTHAGORAS theorem ⇒ Base = √(1 – x²) sec A = Perpendicular/Base = $(\frac{1}{{\SQRT {1 - {x^2}} }})$ tan A = Hypotenuse/Base = $(\frac{x}{{\sqrt {1 - {x^2}} }})$ tan A sec A = $(\frac{x}{{1 - {x^2}}})$

205.

If sin A = x, then the value of tan A sec A1). $\frac{x}{{\sqrt {1 + {x^2}} }}$2). $\sqrt {1 - {x^2}} $3). $\sqrt {\frac{{1 - {x^2}}}{{{x^2}}}}$4). $\frac{x}{{1 - {x^2}}}$

Answer»

sin A = Perpendicular/Hypotenuse = X/1

From PYTHAGORAS theorem ⇒ Base = √(1 – x²)

sec A = Perpendicular/Base = $(\frac{1}{{\SQRT {1 - {x^2}} }})$

tan A = Hypotenuse/Base = $(\frac{x}{{\sqrt {1 - {x^2}} }})$

tan A sec A = $(\frac{x}{{1 - {x^2}}})$

Discussion

206.	Find the value of cot x - tan x1). 2cot 2x2). sec x3). cosec x4). 3cot x
Answer» $(\cot x - \tan x = \;\FRAC{{\cos x}}{{\SIN x}} - \frac{{\sin x}}{{\cos x}})$ = $(\frac{{co{s^2}x - SI{n^2}x}}{{\sin x\;cos\;x}})$ = $(\frac{{2cos2x}}{{2\sin x\;cos\;x}})$ = $(2 \times \frac{{\cos 2x}}{{\sin 2x}} = 2\cot 2x)$

206.

Find the value of cot x - tan x1). 2cot 2x2). sec x3). cosec x4). 3cot x

Answer»

$(\cot x - \tan x = \;\FRAC{{\cos x}}{{\SIN x}} - \frac{{\sin x}}{{\cos x}})$

= $(\frac{{co{s^2}x - SI{n^2}x}}{{\sin x\;cos\;x}})$

= $(\frac{{2cos2x}}{{2\sin x\;cos\;x}})$

= $(2 \times \frac{{\cos 2x}}{{\sin 2x}} = 2\cot 2x)$

Discussion

207.	1). -12). 23). 1/24). 0
Answer» ⇒ LETX = $(\left( {\FRAC{{\sin {\rm{\THETA }} + \sin \phi }}{{\cos {\rm{\theta }} + \cos \phi }} + \frac{{\cos {\rm{\theta }} - \cos \phi }}{{\sin {\rm{\theta }} - \sin \phi }}} \right))$ ⇒ x = [(sin θ + sin ?) (sin θ - sin ?) + (cos θ + cos ?)(cos θ - cos ?)]/(cos θ + cos ?) (sin θ - sin ?) ⇒ x = (sin²θ + cos²θ - sin²? - cos²?)/(cos θ + cos ?) (sin θ - sin ?) ⇒ x = (1 - 1)/(cos θ + cos ?) (sin θ - sin ?) ∴ x = 0

207.

1). -12). 23). 1/24). 0

Answer»

⇒ LETX = $(\left( {\FRAC{{\sin {\rm{\THETA }} + \sin \phi }}{{\cos {\rm{\theta }} + \cos \phi }} + \frac{{\cos {\rm{\theta }} - \cos \phi }}{{\sin {\rm{\theta }} - \sin \phi }}} \right))$

⇒ x = [(sin θ + sin ?) (sin θ - sin ?) + (cos θ + cos ?)(cos θ - cos ?)]/(cos θ + cos ?) (sin θ - sin ?)

⇒ x = (sin²θ + cos²θ - sin²? - cos²?)/(cos θ + cos ?) (sin θ - sin ?)

⇒ x = (1 - 1)/(cos θ + cos ?) (sin θ - sin ?)

∴ x = 0

Discussion

208.	If \(\tan 75^\circ = 2\; + \;\sqrt 3\), then the value of cot215° is –1). \(7\; + \;\sqrt 3 \)2). \(7 - 2\sqrt 3\)3). \(7\; + \;2\sqrt 3\)4). \(7\; + \;4\sqrt 3\)
Answer» tan75° = 2 + √3 tan75° = cot15° = 2 + √3 cot²15° = (2 + √3)² cot²15° = 4 + 3 + 4√3 cot²15° = 7 + 4√3

208.

If $\tan 75^\circ = 2\; + \;\sqrt 3$, then the value of cot215° is –1). $7\; + \;\sqrt 3 $2). $7 - 2\sqrt 3$3). $7\; + \;2\sqrt 3$4). $7\; + \;4\sqrt 3$

Answer»

tan75° = 2 + √3

tan75° = cot15° = 2 + √3

cot²15° = (2 + √3)²

cot²15° = 4 + 3 + 4√3

cot²15° = 7 + 4√3

Discussion

209.	In ΔXYZ measure of angle Y is 90o. If sec X = 17/8, and XY = 0.8 cm, then what is the length (in cm) of side XZ?1). 1.72). 1.53). 24). 2.5
Answer» GIVEN, ∠Y = 90°, sec X = 17/8 ⇒ Sec x = hypotenuse/base = XZ/XY = 17/8 If XY = 0.8 ⇒ 8 unit = 0.8 ⇒ 1 unit = 0.1 ⇒ (XZ) = 17 unit = 1.7 Therefore length of XZ is 1.7 cm

209.

In ΔXYZ measure of angle Y is 90o. If sec X = 17/8, and XY = 0.8 cm, then what is the length (in cm) of side XZ?1). 1.72). 1.53). 24). 2.5

Answer»

GIVEN, ∠Y = 90°, sec X = 17/8

⇒ Sec x = hypotenuse/base = XZ/XY = 17/8

If XY = 0.8

⇒ 8 unit = 0.8

⇒ 1 unit = 0.1

⇒ (XZ) = 17 unit = 1.7

Therefore length of XZ is 1.7 cm

Discussion

210.	If Sin A + Cos A = 0.5, find the value of Sin A.Cos A = ?1). -0.6252). -1.03). 1.454). -0.375
Answer» (SIN A + COS A)² = Sin² A + Cos² A + 2 Sin A Cos A ⇒ (0.5)² = 1 + 2 Sin A Cos A ⇒ 0.25 – 1 = 2 Sin A Cos A ⇒ 2 Sin A Cos A = -0.75 ∴ Sin A Cos A = ?-0.375

210.

If Sin A + Cos A = 0.5, find the value of Sin A.Cos A = ?1). -0.6252). -1.03). 1.454). -0.375

Answer»

(SIN A + COS A)² = Sin² A + Cos² A + 2 Sin A Cos A

⇒ (0.5)² = 1 + 2 Sin A Cos A

⇒ 0.25 – 1 = 2 Sin A Cos A

⇒ 2 Sin A Cos A = -0.75

∴ Sin A Cos A = ?-0.375

Discussion

211.	If sin 13θ = cos 25° (0° < θ < 90°), then the value of θ is1). 3°2). 2°3). 5°4). 12°
Answer» SIN 13θ = COS 25° = sin(90° – 25°) = sin 65° ⇒ 13θ = 65° ⇒ θ = 5°.

211.

If sin 13θ = cos 25° (0° < θ < 90°), then the value of θ is1). 3°2). 2°3). 5°4). 12°

Answer»

SIN 13θ = COS 25° = sin(90° – 25°) = sin 65°

⇒ 13θ = 65°

⇒ θ = 5°.

Discussion

212.	If \(tan\;\theta \; = \;\frac{m}{n}\),then what is \(\frac{{msec\theta- ncosec\theta }}{{msec\theta+ ncosec\theta }}\) equal to?1). \(\frac{{m - n}}{{m + n}}\)2). \(\frac{{{n^2} - {m^2}}}{{{n^2} + {m^2}}}\)3). \(\frac{{{m^2} - {n^2}}}{{{m^2} + {n^2}}}\)4). 1
Answer» Given, TAN θ = m/N ⇒ Hypotenuse = √(m² + n²) $(\begin{array}{l}\frac{{msec\theta- ncosec\theta }}{{msec\theta+ ncosec\theta }}\; = \;\frac{{m\; \times \;\frac{{\sqrt {{m^2} + {n^2}} }}{n} - n\; \times \;\frac{{\sqrt {{m^2} + {n^2}} }}{m}}}{{m\frac{{\sqrt {{m^2} + {n^2}} }}{n} + n\frac{{\sqrt {{m^2} + {n^2}} }}{m}}}\; = \;\frac{{\frac{m}{n} - \frac{n}{m}}}{{\frac{m}{n} + \frac{n}{m}}}\; = \;\frac{{{m^2} - {n^2}}}{{{m^2} + {n^2}}}\\\therefore \;\frac{{msec\theta- ncosec\theta }}{{msec\theta+ ncosec\theta }}\; = \;\frac{{{m^2} - {n^2}}}{{{m^2} + {n^2}}}\end{array})$

212.

If $tan\;\theta \; = \;\frac{m}{n}$,then what is $\frac{{msec\theta- ncosec\theta }}{{msec\theta+ ncosec\theta }}$ equal to?1). $\frac{{m - n}}{{m + n}}$2). $\frac{{{n^2} - {m^2}}}{{{n^2} + {m^2}}}$3). $\frac{{{m^2} - {n^2}}}{{{m^2} + {n^2}}}$4). 1

Answer»

Given,

TAN θ = m/N

⇒ Hypotenuse = √(m² + n²)

$(\begin{array}{l}\frac{{msec\theta- ncosec\theta }}{{msec\theta+ ncosec\theta }}\; = \;\frac{{m\; \times \;\frac{{\sqrt {{m^2} + {n^2}} }}{n} - n\; \times \;\frac{{\sqrt {{m^2} + {n^2}} }}{m}}}{{m\frac{{\sqrt {{m^2} + {n^2}} }}{n} + n\frac{{\sqrt {{m^2} + {n^2}} }}{m}}}\; = \;\frac{{\frac{m}{n} - \frac{n}{m}}}{{\frac{m}{n} + \frac{n}{m}}}\; = \;\frac{{{m^2} - {n^2}}}{{{m^2} + {n^2}}}\\\therefore \;\frac{{msec\theta- ncosec\theta }}{{msec\theta+ ncosec\theta }}\; = \;\frac{{{m^2} - {n^2}}}{{{m^2} + {n^2}}}\end{array})$

Discussion

213.	ΔPQR is right angled at Q. If m∠R = 45°, then find the value of (tan P - 1/2).1). (2 - √3)/√32). (2√3 - √6)/2√23). 1/24). (√6 - 6)/3√3
Answer»

Discussion

214.	Simplify $\sqrt {\left( {1 - {{\sin }^2}\theta } \right) \div \left( {1 - {{\cos }^2}\theta } \right)}$1). cot θ2). tan θ3). sec θ4). cosec θ
Answer» $(\Rightarrow \sqrt {\left( {1 - {{\SIN }^2}\THETA } \RIGHT) \div \left( {1 - {{\cos }^2}\theta } \right)}= \sqrt {\left( {{{\cos }^2}\theta } \right) \div \left( {{{\sin }^2}\theta } \right)}= \sqrt {\left( {{{\cot }^2}\theta } \right)}= \left( {\cot \theta } \right))$

Discussion

215.	\({\left( {1 + {\rm{tanA}}} \right)^2} + {\left( {1{\rm{\;}} - {\rm{\;tanA}}} \right)^2}\) is equal to1). 2cosec2A2). cosec2A3). sec2A4). 2 sec2A
Answer» $(\RIGHTARROW {\rm{\;}}{\LEFT( {1 + {\rm{tanA}}} \right)^2} + {\left( {1{\rm{\;}} - {\rm{\;tanA}}} \right)^2})$ $(\Rightarrow 1 + 2{\rm{tanA}} + {\tan ^2}{\rm{A}} + 1{\rm{\;}} - {\rm{\;}}2{\rm{tanA}} + {\tan ^2}{\rm{A}} = 2 + 2{\tan ^2}{\rm{A\;}})$ $(= 2\left( {1 + {\rm{TA}}{{\rm{N}}^2}{\rm{\;A}}} \right) = {\rm{\;}}2{\sec ^2}{\rm{A}})$

215.

${\left( {1 + {\rm{tanA}}} \right)^2} + {\left( {1{\rm{\;}} - {\rm{\;tanA}}} \right)^2}$ is equal to1). 2cosec2A2). cosec2A3). sec2A4). 2 sec2A

Answer»

$(\RIGHTARROW {\rm{\;}}{\LEFT( {1 + {\rm{tanA}}} \right)^2} + {\left( {1{\rm{\;}} - {\rm{\;tanA}}} \right)^2})$

$(\Rightarrow 1 + 2{\rm{tanA}} + {\tan ^2}{\rm{A}} + 1{\rm{\;}} - {\rm{\;}}2{\rm{tanA}} + {\tan ^2}{\rm{A}} = 2 + 2{\tan ^2}{\rm{A\;}})$

$(= 2\left( {1 + {\rm{TA}}{{\rm{N}}^2}{\rm{\;A}}} \right) = {\rm{\;}}2{\sec ^2}{\rm{A}})$

Discussion

216.	What is the value of Sin 30° + Cos 30°?1). (√6 + 1) /√32). (√3 + 2) /√33). (1 + √3) /24). 5/√3
Answer» SINCE we know that SIN 30 = 1/2 and cos 30 = √3/2, put this in the REQUIRED equation ⇒ Sin 30° + Cos 30° = ½ + √3/2 = (1 + √3) /2 ∴ the value of Sin 30° + Cos 30° is (1 + √3) /2

216.

What is the value of Sin 30° + Cos 30°?1). (√6 + 1) /√32). (√3 + 2) /√33). (1 + √3) /24). 5/√3

Answer»

SINCE we know that SIN 30 = 1/2 and cos 30 = √3/2, put this in the REQUIRED equation

⇒ Sin 30° + Cos 30° = ½ + √3/2 = (1 + √3) /2

∴ the value of Sin 30° + Cos 30° is (1 + √3) /2

Discussion

217.	From the top of a building 75 meters high, the angle of depression of the top and bottom of a tower are observed to be 30° and 60°. The height of the tower in meters is1). 602). 653). 504). 55
Answer»

Discussion

218.	1). 02). 13). 24). √3/2
Answer» tan70° tan20° + SIN²80° + sin² 10° we KNOW that, TAN (π/2 – θ) = cot θ also, sin (π/2 – θ) = cos θ ∴ tan70° tan20° + sin²80° + cos²10° = tan (π/2 – 20°) tan 20° + sin²(π/2 – 10°) + sin²10° = cot20° tan20° + (cos²10° + sin²10°)(? cotθ tanθ = 1 and cos²θ + sin²θ = 1) = 1 + 1 = 2

218.

1). 02). 13). 24). √3/2

Answer»

tan70° tan20° + SIN²80° + sin² 10°

we KNOW that, TAN (π/2 – θ) = cot θ

also, sin (π/2 – θ) = cos θ

∴ tan70° tan20° + sin²80° + cos²10°

= tan (π/2 – 20°) tan 20° + sin²(π/2 – 10°) + sin²10°

= cot20° tan20° + (cos²10° + sin²10°)(? cotθ tanθ = 1 and cos²θ + sin²θ = 1)

= 1 + 1 = 2

Discussion

219.	1). 12/52). 13/53). 5/13 4). 5/12
Answer» Using TRIGONOMETRIC IDENTITIES, sec² θ = 1 + tan² θ ⇒ tan² θ = sec² θ - 1 = 169/144 - 1 = 25/144 ⇒ cot² θ = 1/tan² θ = 144/25 ∴ cot θ = √(144/25) = 12/5

219.

1). 12/52). 13/53). 5/13 4). 5/12

Answer»

Using TRIGONOMETRIC IDENTITIES,

sec² θ = 1 + tan² θ

⇒ tan² θ = sec² θ - 1 = 169/144 - 1 = 25/144

⇒ cot² θ = 1/tan² θ = 144/25

∴ cot θ = √(144/25) = 12/5

Discussion

220.	ΔDEF is right angled at E. If sec D = 25/7, then what is the value of cosec F?1). 7/252). 24/73). 25/74). 7/24
Answer»

Discussion

221.	If \(\frac{{sec\theta\; + \;tan\theta }}{{sec\theta\; - \; tan\theta }} = \frac{5}{3}\), then sinθ is equal to1). 1/42). 1/33). 2/34). 3/4
Answer» $(\frac{{sec\theta\; + \;tan\theta }}{{sec\theta\; - \; tan\theta }} = \frac{5}{3})$ Further solving the above equation we have, $(\frac{{\frac{{1 \;+ \;SIN\theta }}{{cos\theta }}}}{{\frac{{1\; - \; sin\theta }}{{cos\theta }}}} = \frac{5}{3})$ ⇒ 3(1 + sin θ) = 5(1 - sin θ) ⇒ 8sinθ = 2 ⇒ sinθ = 1/4

221.

If $\frac{{sec\theta\; + \;tan\theta }}{{sec\theta\; - \; tan\theta }} = \frac{5}{3}$, then sinθ is equal to1). 1/42). 1/33). 2/34). 3/4

Answer»

$(\frac{{sec\theta\; + \;tan\theta }}{{sec\theta\; - \; tan\theta }} = \frac{5}{3})$

Further solving the above equation we have,

$(\frac{{\frac{{1 \;+ \;SIN\theta }}{{cos\theta }}}}{{\frac{{1\; - \; sin\theta }}{{cos\theta }}}} = \frac{5}{3})$

⇒ 3(1 + sin θ) = 5(1 - sin θ)

⇒ 8sinθ = 2

⇒ sinθ = 1/4

Discussion

222.	1). sinA/22). cosA3). secA4). sinA
Answer»

Discussion

223.	Find the value of 1 – 2 sin2θ + sin4θ.1). sin4θ2). cos4θ3). cosec4θ4). sec4θ
Answer» We KNOW that, sin²x + COS²x = 1 ⇒ sin²θ = 1 – cos²θ Squaring on both sides (? (a + B)² = a² + 2ab + b²) ⇒ sin⁴θ = 1 + cos⁴θ – 2cos²θ Now, 1 – 2 sin²θ + sin⁴θ (? cos 2x = 2cos²x – 1 = 1 – 2sin²x) = 2cos²θ – 1 + 1 + cos⁴θ – 2cos²θ = cos⁴θ ∴ 1 – 2 sin²θ + sin⁴θ = cos⁴θ

223.

Find the value of 1 – 2 sin2θ + sin4θ.1). sin4θ2). cos4θ3). cosec4θ4). sec4θ

Answer»

We KNOW that,

sin²x + COS²x = 1

⇒ sin²θ = 1 – cos²θ

Squaring on both sides

(? (a + B)² = a² + 2ab + b²)

⇒ sin⁴θ = 1 + cos⁴θ – 2cos²θ

Now,

1 – 2 sin²θ + sin⁴θ

(? cos 2x = 2cos²x – 1 = 1 – 2sin²x)

= 2cos²θ – 1 + 1 + cos⁴θ – 2cos²θ

= cos⁴θ

∴ 1 – 2 sin²θ + sin⁴θ = cos⁴θ

Discussion

224.	1). 72). 83). 94). 10
Answer» tan (x² - 8X + 60°) = cot (6x - 5°) tan (x² - 8x + 60°) = tan {90° - (6x - 5°)} x² - 8x + 60° = 90° - 6x + 5 x² - 2x = 35 ∴ x (x -2) = 7× 5 ∴ x = 7

224.

1). 72). 83). 94). 10

Answer»

tan (x² - 8X + 60°) = cot (6x - 5°)

tan (x² - 8x + 60°) = tan {90° - (6x - 5°)}

x² - 8x + 60° = 90° - 6x + 5

x² - 2x = 35

∴ x (x -2) = 7× 5

∴ x = 7

Discussion

225.	1). ½2). 13). 24). 3
Answer» Given, sinA + sin²A = 1 ⇒ sinA = 1 – sin²A = cos²A----(1)(? sin²θ + cos²θ = 1) Squaring on both sides sin²A = cos⁴A----(2) Substitute (1) and (2) in the given expression. cos²A + cos⁴A = sinA + sin²A = 1 ∴ cos²A + cos⁴A = 1

225.

1). ½2). 13). 24). 3

Answer»

Given,

sinA + sin²A = 1

⇒ sinA = 1 – sin²A = cos²A----(1)(? sin²θ + cos²θ = 1)

Squaring on both sides

sin²A = cos⁴A----(2)

Substitute (1) and (2) in the given expression.

cos²A + cos⁴A = sinA + sin²A = 1

∴ cos²A + cos⁴A = 1

Discussion

226.	1). \(\frac{{{a^2} + \sqrt {{a^2} - 1} }}{a}\)2). \(\frac{{{a^2} + \sqrt {1 - {a^2}} }}{a}\)3). \(\frac{{{a^2} - \sqrt {{a^2} - 1} }}{a}\)4). 1
Answer» Sin(90°– a) = cos a Given, cos (10° 7’ 32”) = a Thus, sin (79° 52’ 28”) = a cos²a + sin²a = 1 sin(10° 7’ 32”) = √(1 – a²) tan(10° 7’ 32”) = $(\FRAC{{\sqrt {1 - {a^2}} }}{a})$ sin (79° 52’ 28”) + tan (10° 7’ 32”) $(= a + \frac{{\sqrt {1 - {a^2}} }}{a}\;)$ $(= \frac{{{a^2} + \sqrt {1 - {a^2}} }}{a})$

226.

1). $\frac{{{a^2} + \sqrt {{a^2} - 1} }}{a}$2). $\frac{{{a^2} + \sqrt {1 - {a^2}} }}{a}$3). $\frac{{{a^2} - \sqrt {{a^2} - 1} }}{a}$4). 1

Answer»

Sin(90°– a) = cos a

Given, cos (10° 7’ 32”) = a

Thus, sin (79° 52’ 28”) = a

cos²a + sin²a = 1

sin(10° 7’ 32”) = √(1 – a²)

tan(10° 7’ 32”) = $(\FRAC{{\sqrt {1 - {a^2}} }}{a})$

sin (79° 52’ 28”) + tan (10° 7’ 32”)

$(= a + \frac{{\sqrt {1 - {a^2}} }}{a}\;)$

$(= \frac{{{a^2} + \sqrt {1 - {a^2}} }}{a})$

Discussion

227.	1). 502). $\frac{{200}}{{\sqrt 3 }}$3). 100√34). 100
Answer»

Discussion

228.	What is the value of cosec(-7π/6)?1). -22). 23). 2/√34). -2/√3
Answer» ⇒ COSEC - 7π/6 = -cosec 210° = -cosec (360° - 150°) = - cosec (-150°) [As, the VALUE of cosec REPEATS after 360° INTERVAL] ⇒ cosec (150°) = 2.

228.

What is the value of cosec(-7π/6)?1). -22). 23). 2/√34). -2/√3

Answer»

⇒ COSEC - 7π/6 = -cosec 210° = -cosec (360° - 150°)

= - cosec (-150°) [As, the VALUE of cosec REPEATS after 360° INTERVAL]

⇒ cosec (150°) = 2.

Discussion

229.	If \(\tan \theta = \frac{p}{q}then\frac{{p\sin \theta - q\cos \theta }}{{p\sin \theta + q\cos \theta }}\) is equal to:1). \(\frac{{{P^2} - {q^2}}}{{{p^2} + {q^2}}}\)2). \(\frac{{{P^2} - {q^2}}}{{2pq}}\)3). \(\frac{{2pq}}{{{p^2} - {q^2}}}\)4). \(\frac{{{p^2} + {q^2}}}{{{p^2} - {q^2}}}\)
Answer» Given, $(\FRAC{{\SIN \theta }}{{\COS \theta }} = \frac{p}{q})$ $(\cos \theta = \sin \theta\times \frac{q}{p})$ ⇒ $(? = \frac{{p\sin \theta- q\cos \theta }}{{p\sin \theta + q\cos \theta }})$ ⇒ $(? = \frac{{\left( {\frac{p}{q}\sin \theta - \frac{{\rm{q}}}{{\rm{p}}}\sin \theta } \right)}}{{\left( {\frac{p}{q}\sin \theta + \frac{{\rm{q}}}{{\rm{p}}}\sin \theta } \right)}})$ ⇒ $(? = \frac{{{p^2} - {q^2}}}{{{p^2} + {q^2}}})$

229.

If $\tan \theta = \frac{p}{q}then\frac{{p\sin \theta - q\cos \theta }}{{p\sin \theta + q\cos \theta }}$ is equal to:1). $\frac{{{P^2} - {q^2}}}{{{p^2} + {q^2}}}$2). $\frac{{{P^2} - {q^2}}}{{2pq}}$3). $\frac{{2pq}}{{{p^2} - {q^2}}}$4). $\frac{{{p^2} + {q^2}}}{{{p^2} - {q^2}}}$

Answer»

Given,

$(\FRAC{{\SIN \theta }}{{\COS \theta }} = \frac{p}{q})$

$(\cos \theta = \sin \theta\times \frac{q}{p})$

⇒ $(? = \frac{{p\sin \theta- q\cos \theta }}{{p\sin \theta + q\cos \theta }})$

⇒ $(? = \frac{{\left( {\frac{p}{q}\sin \theta - \frac{{\rm{q}}}{{\rm{p}}}\sin \theta } \right)}}{{\left( {\frac{p}{q}\sin \theta + \frac{{\rm{q}}}{{\rm{p}}}\sin \theta } \right)}})$

⇒ $(? = \frac{{{p^2} - {q^2}}}{{{p^2} + {q^2}}})$

Discussion

230.	If (a² − b²) sinθ + 2ab cosθ = a² + b², then the value of secθ is:1). 1/2 × (a² + b²)2). 1/2ab × (a² – b²)3). 1/2ab × (a² + b²)4). 1/ab × (a² + b²)
Answer» Dividing the whole equation by cosθ, ⇒ (a² − b²)tanθ + 2AB = (a² + b²)secθ (? 1/cosθ = secθ) ⇒ 2ab = (a² + b²)secθ – (a² − b²)tanθ ⇒ 2ab = a² (secθ – tanθ) + b² (secθ + tanθ) Dividing the whole by ab, ⇒ 2 = (a/b)(secθ – tanθ) + (b/a)(secθ + tanθ) ………… (1) Now, ⇒ sec²θ – tan²θ = 1 ⇒ (secθ + tanθ) × (secθ – tanθ) = 1 ⇒ (secθ + tanθ) = 1/(secθ – tanθ) Substitute above in (1), ⇒ 2 = (a/b)(secθ – tanθ) + (b/a) × 1/(secθ – tanθ) Let (a/b)(secθ – tanθ) = Z ⇒ 2 = z + 1/z ⇒ 2z = z² + 1 ⇒ z² – 2z + 1 = 0 ⇒ (z – 1)² = 0 ⇒ z = 1 ⇒ (a/b) (secθ – tanθ) = 1 ⇒ (secθ – tanθ) = b/a ⇒ (secθ + tanθ) = a/b (? it is INVERSE of (sec θ – TAN θ)) Solving above, ⇒ 2secθ = b/a + a/b ⇒ 2secθ = (a² + b²)/ab ∴ secθ = 1/2 ab × (a² + b²)

230.

If (a² − b²) sinθ + 2ab cosθ = a² + b², then the value of secθ is:1). 1/2 × (a² + b²)2). 1/2ab × (a² – b²)3). 1/2ab × (a² + b²)4). 1/ab × (a² + b²)

Answer»

Dividing the whole equation by cosθ,

⇒ (a² − b²)tanθ + 2AB = (a² + b²)secθ (? 1/cosθ = secθ)

⇒ 2ab = (a² + b²)secθ – (a² − b²)tanθ

⇒ 2ab = a² (secθ – tanθ) + b² (secθ + tanθ)

Dividing the whole by ab,

⇒ 2 = (a/b)(secθ – tanθ) + (b/a)(secθ + tanθ) ………… (1)

Now,

⇒ sec²θ – tan²θ = 1

⇒ (secθ + tanθ) × (secθ – tanθ) = 1

⇒ (secθ + tanθ) = 1/(secθ – tanθ)

Substitute above in (1),

⇒ 2 = (a/b)(secθ – tanθ) + (b/a) × 1/(secθ – tanθ)

Let (a/b)(secθ – tanθ) = Z

⇒ 2 = z + 1/z

⇒ 2z = z² + 1

⇒ z² – 2z + 1 = 0

⇒ (z – 1)² = 0

⇒ z = 1

⇒ (a/b) (secθ – tanθ) = 1

⇒ (secθ – tanθ) = b/a

⇒ (secθ + tanθ) = a/b (? it is INVERSE of (sec θ – TAN θ))

Solving above,

⇒ 2secθ = b/a + a/b

⇒ 2secθ = (a² + b²)/ab

∴ secθ = 1/2 ab × (a² + b²)

Discussion

231.	If a person travels from a point L towards east for 12 km and then travels 5 km towards north and reaches a point M, then shortest distance from L to M is:1). 142). 123). 174). 13
Answer»

Discussion

232.	What is the value of (2/√3 + tan45°) ?1). (1 + √6)/√32). (2 + √3)/√33). 4/√34). √3 + 2
Answer» We KNOW that tan45° = 1 Value of (2/√3 + tan45°) = (2/√3) + 1 = (2 + √3)/√3 ∴ REQUIRED ANSWER = (2 + √3)/√3

232.

What is the value of (2/√3 + tan45°) ?1). (1 + √6)/√32). (2 + √3)/√33). 4/√34). √3 + 2

Answer»

We KNOW that tan45° = 1

Value of (2/√3 + tan45°) = (2/√3) + 1 = (2 + √3)/√3

∴ REQUIRED ANSWER = (2 + √3)/√3

Discussion

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