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Correct answer is (B) cos 2β

(b) 16 cos⁴θ – 12 cos²θ + 1 

\(\frac{sin\,5θ}{sin\,θ}\) = \(\frac{sin\,(2\theta+3\theta)}{sin\,\theta}\)

= \(\frac{1}{sin\,θ}\) {sin 2θ cos3θ + cos 2θ sin 3θ}

= \(\frac{1}{sin\,θ}\) \(\big\{\)2 sin θ cos θ (4 cos³ θ - 3 cosθ) + (2 cos² θ -1)(3 sin θ - 4 sin³ θ)\(\big\}\)

(∵ cos3A = 4cos³A – 3cosA, sin 3A = 3sinA – 4sin³A) 

= {2 cos θ (4 cos³θ – 3 cosθ) + (2 cos²θ – 1) (3 – 4 sin²θ)}.  (Note the step) 

= 8 cos⁴θ – 6 cos²θ + 6 cos²θ – 3 – 8 cos²θ sin²θ + 4 sin²θ 

= 8 cos⁴θ – 3 – 8 cos²θ (1 – cos²θ) + 4 (1 – cos²θ)         (∵ sin²θ + cos²θ = 1) 

= 8 cos⁴θ – 3 – 8 cos²θ + 8 cos⁴θ + 4 – 4 cos²θ 

= 16 cos⁴θ – 12 cos²θ + 1.

Number of rotations in 1 min = 1000

So number of rotations in 1 sec = 1000/60 = 50/3

The angle rotated in 1 rotation = 360°

So, the angle rotated in 50/3 rotation = (50/3) x 360° = 6000°

In right angled ∆PQR, 

i. side opposite to ∠P = QR 

ii. side opposite to ∠R = PQ 

iii. side adjacent to ∠P = PQ 

iv. side adjacent to ∠R = QR

In right angled ∆PQR, 

i. side opposite to ∠P = QR 

ii. side opposite to ∠R = PQ 

iii. side adjacent to ∠P = PQ 

iv. side adjacent to ∠R = QR

(i) The value of tan 90° is greater than 1. Therefore, given statement is false. 

(ii)  sec A = \(\frac{12}5\) ⇒ cos A = \(\frac{5}{12}\) as 12 is the hypotenuse largest side. Therefore, given statement is true. 

(iii) cos A is the abbreviation used for cosine of angle A Therefore, given statement is true. 

(iv) cot A is not the product of cot and A. Therefore, given statement is false. 

(v) Since, the hypotenuse is the longest side whereas in sin A = \(\frac{4}{3}\), 3 which is the denominator and cannot be hypotenous.

tan A = \(\frac{15}{8}\) 

∴ sin A = \(\frac{15}{17}\) 

and Cos A =\(\frac{8}{17}\)

∴ Sin A - Cos A = \(\frac{15}{17}\) - \(\frac{8}{17}\) = \(\frac{7}{17}\)

(i) The value of tan 90° is greater than 1. Therefore, given statement is false. 

(ii) sec A = \(\frac{12}{5}\) \(\implies\) cos A = \(\frac{5}{12}\) as 12 is the hypotenuse largest side. Therefore, given statement is true. 

(iii) cos A is the abbreviation used for cosine of angle A Therefore, given statement is true. 

(iv) cot A is not the product of cot and A. Therefore, given statement is false. 

(v) Since, the hypotenuse is the longest side whereas in sin A =\(\frac{4}{3}\) , 3 which is the denominator and cannot be hypotenous.

We know that

cos² θ + sin² θ = 1

⇒ cos² 50° + sin² 50° = 1

⇒ cos ² 50° = 1 – sin² 50°

⇒ cos 50° = √(1 – sin² 50°)

And Given that sin 50° = a

⇒ cos 50° = √(1 – a² )

sin(3 × 10°) = sin(30°) = \(\frac{1}{2}\)

⇒ 3 sin10° - 4 sin³10° = \(\frac{1}{2}\)

⇒ 8 sin³10° - 6 sin10° + 1 = 0

After solving this polynomial in sin10°, we get

⇒ sin10° = 0.17365

sin70° = sin(10° + 60°)

= sin10° cos60° + cos10° sin60°

= 0.17365 × \(\frac{1}{2}\) + 0.9848 × \(\frac{\sqrt3}{2}\)

= 0.086825 + 0.852868

= 0.93693

\(\frac{1}{sin 10°}\) - 4 sin70° = \(\frac{1}{0.17365} - 4(0.93693)\)

= 5.75871 - 3.74772

= 2.01099 (2.011 Approx)

False

Justification:

L.H.S = (tan θ+2) (2 tan θ+1)

= 2 tan² θ + tan θ + 4 tan θ + 2

= 2 tan²θ+5 tan θ+2

Since, sec² θ – tan² θ = 1, we get, tan²θ = sec² θ-1

= 2(sec² θ-1) +5 tan θ+2

= 2 sec² θ-2+5 tan θ+2

= 5 tan θ+ 2 sec² θ ≠R.H.S

∴, L.H.S ≠ R.H.S

(sin 72° + cos 18°)( sin 72° - cos 18°)

Using the identity , (a - b)(a + b) = a² – b²

= (sin 72°)² - (cos 18°)²

= {cos(90° - 72°)}² - (cos 18°)² [∵ Sin θ = cos (90° - θ)]

= (cos 18°)² - (cos 18°)²

= 0

We know that

cos² θ + sin² θ = 1

⇒ sin² θ = 1 – cos² θ

⇒ sin θ = √(1 – cos² θ)

And Given that cos θ = x

⇒ sin θ = √(1– x²)

Correct option is: D)  \(\frac 54\)

\(\because\)  -1 \(\leq\) sin x \(\leq\) 1

Any value which is greater than 1 or less than -1 is not a possible value of sin x.

\(\because\) \(\frac 54\) > 1

\(\therefore\) \(\frac 54\) is not a possible value of sin x.

Correct option is: D) \(\frac{5}{4}\)

Correct option is: B) \(\frac{7}{24}\)

We have sin A = \(\frac{24}{25}\)

\(\therefore\) \(cos^2A = 1-sin^2A = 1-(\frac {24}{25})^2 = 1- \frac {576}{625} = \frac {625 -576}{625}\)

= \(\frac {49}{625} = \frac {7^2}{25^2} = (\frac 7{25})^2\)

\(\therefore\) cos A = \(\frac 7 {25}\) 

\(\therefore\) cot A = \(\frac {cos\, A}{sin \, A} = \frac {\frac 7{25}}{\frac {24}{25}} = \frac 7{24}\)

Correct option is: B) \(\frac{7}{24}\)

cosec² 72° − tan² 18° = 1

LHS = cosec² 72° – tan² 18°

= cosec² 72° – tan² (90 – 72)°

= cosec² 72° – cot² 72°

= 1 = RHS

We know that

cos² θ + sin² θ = 1

⇒ cos² 40° + sin² 40° = 1

⇒ sin² 40° = 1 – cos² 40°

⇒ sin 40° = √(1 – cos² 40°)

And Given that cos 40° = p

⇒ sin 40° = √(1– p²)

Correct option is: B) Tan 90°

\(\because\) (a) tan \(0^\circ\) = 0

(b) tan \(90^\circ\) = \(\theta\) (Not defined)

(c) cot \(90^\circ\) = 0

(d) sec \(0^\circ\) = 1

Correct option is: B) Tan 90°

Correct option is: D) Sin 60°

\(\frac{1-Tan^2 45^\circ}{1+Tan^2 45^\circ}\) = \(\frac {1-1^2}{1+1^2} = \frac {1-1}{1+1} = \frac 02 = 0\) (\(\because\) tan \(45^\circ\) = 1)

\(\because\) sin \(0^\circ\) = 0, cos \(90^\circ\)  = 0

and sin \(0^\circ\). cos \(90^\circ\)  = 0 \(\times\) 0 = 0

But sin \(60^\circ\) = \(\frac {\sqrt3}{2} \neq 0\)

\(\therefore\) \(\frac{1-Tan^2 45^\circ}{1+Tan^2 45^\circ}\) can not be expressed as sin \(60^\circ\).

Correct option is: D) Sin 60°

We know that, 

sin (90 – θ) = cos θ 

So, the given can be expressed as 

(sin 72° + cos 18°) (sin (90 – 18)° – cos 18°) 

= (sin 72° + cos 18°) (cos 18° – cos 18°) 

= (sin 72° + cos 18°) x 0 

= 0

We have, 

cosec 31° – sec 59° 

Since, cosec (90 – θ) = cos θ 

So, 

cosec 31° – sec 59° 

= cosec (90° – 59°) – sec 59° 

= sec 59° – sec 59° = 0 

Thus, 

cosec 31° – sec 59° = 0

Taking the L.H.S, 

sinθ sin (90° – θ) – cos θ cos (90° – θ) 

= sin θ cos θ – cos θ sin θ [∵ sin (90 – θ) = cos θ and cos (90 – θ) = sin θ] 

= 0

(sin θ – cosec θ)² + (cos θ – sec θ)² 

= sin² θ + cosec² θ – 2 sinθ . cosecθ . + cos² θ + sec² θ – 2 cosθ . sec θ 

= (sin² θ + cos² θ) + cosec² θ + sec² θ – 2 – 2 

= 1 + (1 + cot² θ) + (1 + tan² θ) – 2 – 2

= cot² θ + tan² θ + 3 – 4 

= cot² θ + tan² θ – 1

sin θ < 0 

sin θ is negative in 3rd and 4th quadrants, cos θ < 0 

cos θ is negative in 2nd and 3rd quadrants. 

∴ θ lies in the 3rd quadrant.

tan θ < 0 tan θ is negative in 2nd and 4th quadrants, sec θ > 0 

sec θ is positive in 1st and 4th quadrants. 

∴ θ lies in the 4th quadrant.

Correct option is: B) \(\frac{4}{5}\)

We have sin A = \(\frac 35\)

Now \(cos^2\, A = 1 -sin^2A = 1 -(\frac 35)^2 = 1 - \frac 9{25} \)

\(= \frac {25-9}{25} = \frac {16}{25} = (\frac 45)^2\)

\(\therefore\) cos A = \(\frac{4}{5}\)

Correct option is: B) \(\frac{4}{5}\)

LHS = cos81° – sin9°

= cos(90° - 9°)- sin9°

= sin9° – sin9°

= 0

= RHS

(i) LHS = cos 81° - sin 9°

= cos(90° - 9°) - sin 9°

= sin 9° - sin 9°

= 0 

= RHS

(ii) LHS = tan 71° - cot 19°

= tan(90° - 19°) - cot 19°

= cot 19° - cot 19°

= 0 

= RHS

(iii) LHS = cosec 80° - sec 10°

= cosec (90° - 10°) - sec 10°

= sec 10° - sec 10°

= 0 

= RHS

(iv) LHS = cosec² 72° - tan² 18°

= cosec² (90° - 18°) - tan² 18°

= sec² 18° - tan² 18°

= 1

= RHS

(v) LHS = cos² 75° - cos² 15°

= cos² (90° - 15°) + cos² 15°

= sin² 15° + cos² 15°

= 1

= RHS

 (i) \(\frac{cos35°}{sin55°}\)

=  \(\frac{cos(90°-55°)}{sin55°}\)

= \(\frac{sin55°}{sin55°}\) [∵ sin(90 - θ) = cos θ]

= 1

(ii) \(\frac{cosec42°}{sec48°}\)

= \(\frac{cosec(90°-48°)}{sec48°}\)

= \(\frac{sec48°}{sec48°}\) [∵ sec(90 - θ) = cosec θ]

= 1

(iii) \(\frac{cot38°}{tan52°}\)

= \(\frac{cot(90°-52°)}{tan52°}\)

= \(\frac{tan52°}{tan52°}\) [∵ tan(90 - θ) = cot θ]

= 1

(i) \(\frac{sin16°}{cos74°}\)

= \(\frac{sin(90°-74°)}{cos74°}\)

= \(\frac{cos74°}{cos74°}\) [∵ sin(90 - θ) = cos θ]

= 1

(ii) \(\frac{sec11°}{cosec79°}\)

= \(\frac{sec(90°- 79°)}{cosec79°}\)

= \(\frac{cosec79°}{cosec79°}\) [∵ sec(90 - θ) = cosec θ]

= 1

(iii) \(\frac{tan27°}{cot63°}\)

 = \(\frac{tan(90°- 63°)}{cot63°}\)

= \(\frac{cos63°}{cos63°}\) [∵ tan(90 - θ) = cot θ]

= 1

(i) LHS = \(\frac{2sin68^\circ}{cos22^\circ}-\frac{2cot15^\circ}{5tan75^\circ}- \frac{3{tan45^\circ}{tan20^\circ}{tan40^\circ}{tan50^\circ}{tan70^\circ}}5\)

= \(\frac{2sin68^\circ}{sin(90^\circ-22^\circ)}-\frac{2cot15^\circ}{5tan75^\circ}- \frac{3\times1\times{cot(90^\circ-20^\circ)}\times{cot(90^\circ-40^\circ)}\times{tan50^\circ}\times{tan70^\circ}}5\)

= \(\frac{2sin68^\circ}{sin68^\circ}-\frac{2cot15^\circ}{5cot15^\circ}- \frac{3\times{cot70^\circ}{tan50^\circ}tan70^\circ}5\)

= 2 - \(\frac{2}5\)- \(\frac{3\times{\frac{1}{tan70^\circ}}\times{\frac{1}{tan50^\circ}}\times{tan50^\circ}\times{tan70^\circ}}5\)

= 2 - \(\frac{2}5\)- \(\frac{3}5\)

= \(\frac{10-2-3}5\)

= \(\frac{5}5\)

= 1

= RHS

(ii) LHS =  \(\frac{7cos55^\circ}{3sin35^\circ}-\frac{4(cos70^\circ{cosec20^\circ})}{3(tan5​​°tan25^\circ{tan45^\circ}tan65^\circ{tan85^\circ})}\)

= \(\frac{7cos55^\circ}{3cos(90^\circ-35^\circ)}-\frac{4(sin(90^\circ-70^\circ)cosec 20^\circ}{{3cot(90^\circ- 5^\circ)}\times{cot(90^\circ-25^\circ)}\times1\times{tan65^\circ}\times{tan85^\circ}}\)

= \(\frac{7cos55^\circ}{3cos55^\circ}\) - \(\frac{{4(sin20^\circ}{cosec20^\circ})}{{3({cot85^\circ}{cot65^\circ}{tan65^\circ}{tan85^\circ}}}\)

= \(\frac{7}3\) - \(\frac{{4(sin20^\circ}\times{\frac{1}{tan20^\circ}})}{{3({\frac{1}{tan85^\circ}}\times{\frac{1}{tan65^\circ}}{tan65^\circ}{tan85^\circ)}}}\) 

= \(\frac{7}3\) - \(\frac{4}3\)

= \(\frac{3}3\)

= 1

= RHS

We know: cot A = tan (90° – A)

So, cot 25° = tan (90° – 25°) = tan 65°

i.e., tan 65°/cot 25° = tan 65°/ tan 65°=1

⇒ tan 5° x tan 30° x 4 tan 85° 

As, 

tan (90 – θ) = cot θ 

Therefore, 

⇒ tan (90 – 85) x tan 30° x 4 tan 85° 

⇒ cot 85° × tan 85° × 4 × tan 30° 

⇒ 1/tan 85 × tan 85° × 4 × tan 30° 

As we know, 

tan 30° = 1/√3 

⇒ 4 × tan 30° 

⇒ 4 × (1/√3) = 4/√3

L.H.S.

= cosec (65° + θ) - sec (25° - θ) - tan (55° - θ) + cot (35° + θ)

= cosec (65° + θ) - sec (90° – (65° + θ)) - tan (90° – (35° + θ)) + cot (35° + θ)

= cosec (65° + θ) - cosec (65° + θ)) - cot (35° + θ) + cot (35° + θ)

= 0

= R.H.S.

Correct answer = (b) 1/√3

We have

tan 5° tan 25° tan 30° tan 65° tan 85°

= tan 5° tan 25° tan 30° tan(90° − 25°) tan(90° − 5°) 

= tan 5° tan 25° × 1/√3 × cot 25° cot 5° [∵ tan(90° − θ) = cot θ and tan 30° = 1/√3 ] 

= 1/√3

tan 35° = tan (90° - 55°) = cot 55°
tan 40° = tan (90° - 50°) = cot 50°

Cot 55 tan 55∙ cot 50 tan 50
=1 ∙ 1  

= 1

We have, 

[∵ cot (90 – θ) = tan θ and tan (90 – θ) = cot θ]

\(\frac{tan\, 35°}{cot\, 55°} + \frac{cot\, 78°}{tan\, 12°} - 1\)

= \(\frac{tan\, (90°-\, 35°)}{cot\, 55°} + \frac{cot\,(90° -\, 12°)}{tan\, 12°} - 1\)

= \(\frac{cot\, 55°}{cot\, 55°} + \frac{tan\, 12°}{tan\, 12°} - 1\)

= 1 + 1 – 1 

= 1

Given: sin θ +2 cos θ = 1

Squaring on both sides,

(sin θ +2 cos θ)² = 1

⇒ sin² θ + 4 cos² θ + 4sin θcos θ = 1

Since, sin² θ = 1 – cos² θ and cos² θ = 1 – sin² θ

⇒ (1 – cos² θ) + 4(1 – sin² θ) + 4sin θcos θ = 1

⇒ 1 – cos² θ + 4 – 4 sin² θ + 4sin θcos θ = 1

⇒ – 4 sin² θ – cos² θ + 4sin θcos θ = – 4

⇒ 4 sin² θ + cos² θ – 4sin θcos θ = 4

We know that,

a²+ b² – 2ab = ( a – b)²

So, we get,

(2sin θ – cos θ)² = 4

⇒ 2sin θ – cos θ = 2

Hence proved.

L.H.S. = tan 48° tan 16° tan 42° tan 74° 

= tan 48°. tan 16° . tan(90° – 48°) . tan(90° – 16°) 

= tan 48° . tan 16° . cot 48° . cot 16° [∵ tan (90 – θ) = cot θ] 

= tan 48° . tan 16° . \(\frac{1}{tan \,48^∘}\) . \(\frac{1}{tan \,16^∘}\) [∵ cot θ = \(\frac{1}{tan \,θ}\)]

= 1 = R.H.S. 

∴ L.H.S. = R.H.S.

Correct option is: B) 1

Given that cosec \(\theta\) = 2

and cot \(\theta\) = \(\sqrt {3P}\)

= \(cot^2 \theta\) = 3P

= \(cosec^2 \theta -1 \) = 3P ( \(\because\) \(cot^2 \theta\) = \(cosec^2 \theta -1 \))

= 4-1 = 3P (\(\because\) cosec \(\theta\) = 2)

= 3P = 3

= P = \(\frac 33\) = 1

Correct option is: B) 1

sec²θ (1 + sinθ)(1 - sinθ)

= sec² θ(1 − sin² θ) 

= 1/cos² θ × cos² θ 

= 1

sin²θ cos²θ(1 + tan²θ)(1 + cot²θ)

= sin² θ cos² θ sec² θ cosec² θ 

= sin² θ × cos² θ × 1/cos² θ × 1/sin² θ 

= 1

cosec²θ(1 + cosθ)(1 - cosθ)

= cosec²θ(1 − cos² θ) 

= 1/sin² θ × sin² θ 

= 1

(c) \(\bigg]\frac{π}{2},\frac{2π}3\bigg[\)

sin θ = \(\frac12\) ⇒ sin θ = sin \(\fracπ6\) ⇒ θ = \(\fracπ6\)            

(∵ θ and φ are acute angles lying in the first quadrant)            …(i) 

Now cos ϕ = \(\frac13\) ⇒ 0 < cos ϕ < \(\frac12\) ⇒ cos \(\fracπ2\) < cos ϕ < cos \(\fracπ3\) ⇒ \(\fracπ2\) < ϕ < \(\fracπ3\)         …(ii) 

∴ From (i) and (ii) \(\fracπ2\) + \(\fracπ6\) < θ + ϕ < \(\fracπ3\) + \(\fracπ6\)

⇒ \(\fracπ2\) < θ + ϕ < \(\frac{2π}3\) ⇒ θ + ϕ lies in the open interval \(\bigg]\frac{π}{2},\frac{2π}3\bigg[\). 

Hence (c) is the correct option.

4 sin²x + 4 cos \(x\) = 1 

⇒ 4 sin²x + 4 cos \(x\) – 1 = 0 ⇒ 4 (1 – cos²x) + 4 cos \(x\) – 1 = 0 

⇒ – 4 cos²x + 4 cos x + 3 = 0 ⇒ 4 cos²x – 4 cos \(x\) – 3 = 0 

⇒ 4 cos²x – 6 cos \(x\) + 2 cos \(x\) – 3 = 0 ⇒ 2 cos \(x\) (2 cos \(x\) – 3) + 1 (2 cos \(x\) – 3) = 0 

⇒ (2 cos \(x\) + 1) (2 cos \(x\) – 3) = 0 

⇒ cos \(x\) = \(-\frac12\) and cos \(x\) = \(\frac32\) (not possible) 

Now cos \(x\) = \(-\frac12\) 

Since A lies in the third quadrant and cos A = \(-\frac12\), therefore, 

cos A = cos (180º + 60º) = cos 240º ⇒ A = 240º.

(b) 3

Given, sec θ = \(\frac{13}{5}\) ⇒ sec² θ = \(\frac{169}{25}\)

⇒ 1 + tan²θ = \(\frac{169}{25}\)                 (∵ sec²θ – tan²θ = 1) 

⇒ tan² θ = \(\frac{169}{25}\) - 1 = \(\frac{144}{25}\) ⇒ tan θ = \(\frac{12}{5}\)

∴ \(\frac{2\,sin\,\theta-3\,cos\,\theta}{4\,\sin\,\theta-9\,cos\,\theta}\) = \(\frac{\frac{2\,sin\,\theta}{cos\,\theta}-3}{\frac{4\,sin\,\theta}{cos\,\theta}-9}\)

(On dividing each term of numerator and denominator by cos θ)

\(= \frac{2\,tan\,\theta-3}{4\,tan\,\theta-9}=\frac{2\times\frac{12}{5}-3}{4\times\frac{12}{5}-9}\)

= \(\frac{24-15}{48-45}=\frac93=3.\)

√3 cosec 20° – sec 20°

= \(\frac{\sqrt3}{sin\,20°}\) - \(\frac{1}{cos\,20°}\) = \(\frac{​​\sqrt3\,cos\,20°-sin\,20°}{sin20°\,cos20°}\)

= \(\frac{​​4}{2\,sin20°\,cos20°}\bigg(\frac{\sqrt3}{2}cos\,20°-\frac12sin\,20°\bigg)\) (Multiplying numerator and denominator by 4)

= \(\frac{4}{sin\,40°}\) (sin 60° cos 20° – cos 60° sin 20°)    (∵ sin 2θ = 2 sin θ cos θ)

= \(\frac{4}{sin\,40°}\) (sin (60° – 20°))         (∵ sin (A – B) = sin A cos B – cos A sin B)

= \(\frac{4}{sin\,40°}\). sin 40° = 4.

Given sin 56° = x

We know that sin θ = cos (90° - θ)

Here, θ = 56°

⇒ cos (90° - 56°) = x

⇒ cos 34° = x

Given sin 37° = a

We know that sin θ = cos (90° - θ)

Here, θ = 37°

⇒ cos (90° - 37°) = a

⇒ cos 53° = a