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(a) \(\frac{1}{\sqrt{3}}\)

\(\frac{3tan10°-tan^310°}{1-3tan^210°}\) = tan(3 x 10°) [∵ tan 3A = \(\frac{3tanA-tan^3A}{1-3tan^2A}\)]

= tan 30°

= \(\frac{1}{\sqrt{3}}\)

\(\frac{5\,sin\,75°\,sin\,77°+2\,cos\,13°cos\,15°}{cos\,15°sin77°}\) - \(\frac{7\,sin\,81°}{cos\,90}\)

= \(\frac{5\,sin\,(90°-15°)\,sin\,77°+2\,cos\,(90°-77°)cos\,15°}{cos\,15°sin\,77°}\) - \(\frac{7\,sin\,(90°-9°)}{cos\,9°}\)

= \(\frac{5\,cos\,15°\,sin\,77°+2\,sin\,77°cos\,15°}{cos\,15°sin\,77°}\) - \(\frac{7\,sin\,9°}{cos\,9°}\)

= \(\frac{5\,cos\,15°\,sin\,77°}{cos\,15°sin\,77°}\) - \(\frac{7\,sin\,9°}{cos\,9°}\) = 7 - 7 = 0.

(b) \(-\frac{\sqrt{3}}{2}\)

sin(-420°) = -sin(420°) [∵ sin(-θ) = -sin θ] 

= -sin(360° + 60°) 

= -sin 60° 

= \(-\frac{\sqrt{3}}{2}\)

(b) cos² 75° + cos² 70° + ........ + cos² 15°

sin² 15° + sin² 20° + sin² 25° + ........ + sin² 75° 

= sin² (90° – 75°) + sin² (90° – 70°) + ... + sin² (90° – 15°) 

= cos² 75° + cos² 70° + ........ + cos² 15°

(b)  -1 

sin 120° cos 150° – cos 240° sin 330° 

= sin (180° – 60°) cos (180° – 30°) – cos (180° + 60°) sin (360° – 30°)

= (sin 60°) . (– cos 30°) – (– cos 60°) (– sin 30°)

= \(\frac{\sqrt3}{2}\times-\frac{\sqrt3}{2}-\big(\frac{-1}{2}\big)\big(\frac{-1}{2}\big)\) = \(-\frac34-\frac14=-\frac44=-1.\)

520° =360° + 160° 

∴ 520° and 160° are co-terminal angles. 

Since 90° < 160° < 180°, 

160° lies in the 2nd quadrant.

∴ 520° lies in the 2nd quadrant, 

∴ cosec 520° is positive.

2cos 35° cos75° 

= cos(35° + 75°) + cos (35° – 75°) 

= cos 110° + cos (-40)° 

= cos 110° + cos 40° … [∵ cos(-θ) = cos θ]

2sin 4x cos 2x 

= sin(4x + 2x ) + sin (4x – 2x) 

= sin 6x + sin 2x

2 sin 2π/3 cos π/2 = sin (\(\frac{2π}{3}+\frac{π}{2} \)) sin (\(\frac{2π}{3}-\frac{π}{2} \))

= sin \(\frac{7π}{6}+sin\, \frac{π}{6} \)

cos⁵ θ = cos(2θ + 3θ) = cos 2θ cos 3θ – sin 2θ sin 3θ 

= (2 cos² θ – 1) (4 cos³ θ – 3 cos θ) – 2 sin θ cos θ (3 sin θ – 4 sin³ θ)

= 8 cos⁵ θ – 6 cos³ θ – 4 cos³ θ + 3 cos θ – 6 sin² θ cos θ + 8 cos θ sin⁴ θ 

= 8 cos⁵ θ – 6 cos³ θ – 4 cos³ θ + 3 cos θ – 6(1 – cos² θ) cos θ + 8 cos θ (1 – cos² θ)² 

= 8 cos⁵ θ – 6 cos³ θ – 4 cos³ θ + 3 cos θ – 6 cos θ + 6 cos³ θ + 8 cos 0(1+ cos⁴ θ – 2 cos² θ) 

= 8 cos⁵ θ – 6 cos³ θ – 4 cos³ θ + 3 cos θ – 6 cos θ + 6 cos³ θ + 8 cos θ + 8 cos⁵ θ – 16 cos³ θ 

= 16 cos³ θ – 20 cos³ θ + 5 cos θ = RHS 

Correct option is (B) 1

tan1° tan2° tan3° … tan89° 

= (tan 1° tan 89°) (tan 2° tan 88°) 

…(tan 44° tan 46°) tan 45° 

= (tan 1 ° cot 1 °) (tan 2° cot 2°) 

…(tan 44° cot 44°) . tan 45° 

…tan(∵ 90° – θ) = cot θ] 

= 1 x 1 x 1 x … x 1 x tan 45° = 1

Correct answer is (B) 1

Explanation :

tan1° tan2° tan3° … tan89° 

= (tan 1° tan 89°) (tan 2° tan 88°) 

…(tan 44° tan 46°) tan 45° 

= (tan 1 ° cot 1 °) (tan 2° cot 2°) 

…(tan 44° cot 44°) . tan 45° 

…tan(∵ 90° – θ) = cot θ] 

= 1 x 1 x 1 x … x 1 x tan 45° = 1

A + B = 45° 

tan (A + B) = tan 45° = 1

(i.e) (tan A + tan B)/(1 - tan A tan B) = 1

tan A + tan B = 1 - tan A tan B ... (1) 

Now LHS = (1 + tan A) (1 + tan B) 

= tan A + tan B + tan A tan B + 1 

= (1 – tan A tan B) + (tan A tan B + 1) from (1) 

= 2 = RHS

Given A + B = 45° 

tan (A + B) = tan 45°

\(\frac{tanA+tanB}{1-tanAtanB}\) = 1

tan A + tan B = 1 – tan A.tan B 

tan A + tan B + tan A tan B = 1

Add 1 on both sides we get, 

(1 + tan A) + tan B + tan A tan B = 2 

1(1+ tan A) + tan B (1 + tan A) = 2 

(1 + tan A) (1 + tan B) = 2 … (1) 

Put A = B = 22\(\frac{1}{2}\) in (1) we get 

(1 + tan 22\(\frac{1}{2}\)) (1 + tan 22\(\frac{1}{2}\)) = 2 

⇒ (1 + tan 22\(\frac{1}{2}\))² = 2 

⇒ 1 + tan 22\(\frac{1}{2}\) = ±√2 

⇒ tan 22\(\frac{1}{2}\) = ±√2 – 1 

Since 22\(\frac{1}{2}\) is acute, tan 22\(\frac{1}{2}\) is positive and 

Therefore tan 22\(\frac{1}{2}\) = √2 – 1 

2 sin 30° + cos 0° + 3 sin 90°

2 sin 30° + cos0° + 3 sin 90° = 2 (1/2 ) + 1 + 3(1) 

= 1 + 1 + 3 

∴ 2 sin 30° + cos 0° + 3 sin 90° = 5

Correct answer is  (B) 2

Taking LHS = cos 27° + sin51^o

We know that

cos θ = sin (90° - θ)

Here, θ = 27°

⇒ sin (90° - 27°)+ sin 51°

⇒ sin 63°+ sin 51°

We also know that

Sin θ = cos (90° - θ)

Here, θ = 51°

⇒ sin 63°+ cos (90° - 51°)

⇒ sin 63°+ cos 39° = RHS

Hence Proved

Taking LHS

= sin 9° sin 27° sin 63° sin 81°

= cos (90° - 9°) cos (90° - 27°) cos (90° – 63°) cos (90°- 81°)

= cos 81° cos 63° cos 27° cos 9°

Or cos 9° cos 27° cos 63° cos 81° = RHS

Hence Proved

cos60° cos45° - sin60° sin 45°

= \(\frac{1}2\) x \(\frac{1}{\sqrt{2}}\) - \(\frac{\sqrt3}2\) x \(\frac{1}{\sqrt{2}}\)

= \(\frac{1-\sqrt3}{2\sqrt2}\)

Correct answer is (b) 1/4

(Sin² 30° + 4 cot² 45° − sec² 60°) 

= [( 1/2 )² + 4×(1)² − (2)²] 

= (1/4 + 4 − 4) = 1/4

We know that

Sin (90°) = 1

Cos (0°) = 1

Tan(0°) = 0

Tan(45°) = 1

Now putting the value, we get

= 1 – 1 + 0 + 1

= 1

\(x\) = y cos 120° = z cos 240° 

⇒ \(x\) = y cos (180° – 60°) = z cos (180° + 60°) 

⇒ \(x\) = – y cos 60° = – z cos 60° (∵ cos (180° – θ) = – cos θ = cos (180° + θ)) 

⇒ \(x\) = \(-\frac12\) y = \(-\frac{z}2\) ⇒ 2\(x\) = – y = – z 

⇒ \(\frac{x}{\frac12}\) = \(\frac{y}{(-1)}\) = \(\frac{z}{(-1)}\) = k 

⇒ \(x\) = \(\frac{k}{2}\) , y = – k, z = – k 

∴ xy + yz + zx = \(\big(\frac{k}{2}\big)\) (–k) + (–k) (–k) + (–k) \(\big(\frac{k}{2}\big)\) = \(\frac{-k^2}{2}\) + k² – \(\frac{k^2}{2}\) = 0.

Correct option is: A) \(\frac{1}{2}\)

\(\frac{1-tan^2 30^\circ}{1+tan^2 30^\circ} = \frac {1-(\frac 1{\sqrt3})^2}{1+ (\frac 1{\sqrt3})^2}\) ( \(\because\) tan \(30^\circ\) = \(\frac 1{\sqrt3}\))

= \(\frac {1-\frac 13}{1+ \frac 13} = \frac {\frac {3-1}3}{\frac {3+1}3} = \frac 24 = \frac 12\)

Correct option is: A) \(\frac{1}{2}\)

Correct option is: A) \(cos^2 \theta – sin^2 \theta\)

\(cos^4 \theta – sin^4 \theta \) = \((cos^2\theta - sin^2\theta)(cos^2\theta+ sin^2\theta)\) (\(\because\) \(a^2-b^2 = (a-b) (a+b)\))

= \(\cos^2\theta - sin^2\theta \) (\(\because\) \(\cos^2\theta - sin^2\theta \) = 1).

Correct option is: A) cos² θ – sin² θ