Answer:
1. Answer: (b) sin2 (x + y)
Explanation: cos2x + cos2y – 2co x × cosy × cos(x + y)
{since cos(x + y) = cosx x cos y – sin x × sin y }
= cos2x + cos2y – 2cosx × cosy × (cosx × cosy – sinx × siny)
= cos2x + cos2y – 2cos2x × cos2y + 2cosx × cosy × sinx × siny
= cos2x + cos2y – cos2x × cos2y – cos2x × cos2y + 2cosx × cosy × sinx × siny
= (cos2x – cos2 x × cos2 y) + (cos2y – cos2x × cos2y) + 2cosx × cosy × sinx × siny
= cos2x(1- cos2y) + cos2 y(1 – cos2x) + 2cosx × cosy × sinx × siny
= sin2y × cos2x + sin2x × cos2y + 2cosx × cosy × sinx × siny (since sin2x + cos2x = 1 )
= sin2x × cos2y + sin2y × cos2x + 2cosx × cosy × sinx × siny
= (sinx × cosy)2 + (siny × cosx)2 + 2cosx × cosy × sinx × siny
= (sinx × cosy + siny × cosx)2
= {sin (x + y)}2
= sin2 (x + y)
2. Answer: (d) a2 + b2 – c2
Explanation: a × cosx + b × sinx)2 + (a × sinx – b × cosx)2 = a2 + b2
⇒ c2+ (a × sinx – b × cosx)2 = a2 + b2
⇒ (a × sinx – b × cosx)2 = a2 + b2 – c2
3. Answer: (c) -1
Explanation: cos 5π = cos (π + 4π)
cos ( 5 × 180° )
= cos 900°
= cos ( 10 × 90° + 0° )
= - cos 0°
= - 1
4. Answer: (c) 1
Explanation: Given cosec A (sin B cos C + cos B sin C)
= cosecA × sin(B+C)
= cosecA × sin(180 – A)
= cosecA × sin A
= cosecA × 1/cosec A
= 1
5. Answer: (c) -1
Explanation: cos 180 = cos(90° + 90°)
= -sin 90°
= -1 {sin90° = 1}
cos180° = -1
6. Answer: (b) 90°
Explanation: Let the lengths of the sides if ∆ABC be a, b and c
Perimeter of the triangle = 2s = a + b + c = 6(sinA + sinB + sinC)/3
⇒ (sinA + sinB + sinC) = ( a + b + c)/2
⇒ (sinA + sinB + sinC)/( a + b + c) = 1/2
From sin formula,Using
sinA/a = sinB/b = sinC/c = (sinA + sinB + sinC)/(a + b + c) = 1/2
Now, sinB/b = 1/2
Given b = 2
So, sinB/2 = 1/2
⇒ sinB = 1
⇒ B = π/2
7. Answer: (b) 45
Explanation: Given, 3×tan(x – 15) = tan(x + 15)
⇒ tan(x + 15)/tan(x – 15) = 3/1
⇒ {tan(x + 15) + tan(x – 15)}/{tan(x + 15) – tan(x – 15)} = (3 + 1)/(3 – 1)
⇒ {tan(x + 15) + tan(x – 15)}/{tan(x + 15) – tan(x – 15)} = 4/2
⇒ {tan(x + 15) + tan(x – 15)}/{tan(x + 15) – tan(x – 15)} = 2
⇒ sin(x + 15 + x – 15)/sin(x + 15 – x + 15) = 2
⇒ sin 2x/sin 30 = 2
⇒ sin 2x/(1/2) = 2
⇒ 2 × sin 2x = 2
⇒ sin 2x = 1
⇒ sin 2x = sin 90
⇒ 2x = 90
⇒ x = 45
8. Answer: (c) 2π/3
Explanation: Given, the sides of a triangle are 13, 7, 8
Since greatest side has greatest angle,
Now Cos A = (b2 + c2 – a2)/2bc
⇒ Cos A = (72 + 88 – 132)/(2×7×8)
⇒ Cos A = (49 + 64 – 169)/(2×7×8)
⇒ Cos A = (113 – 169)/(2×7×8)
⇒ Cos A = -56/(2×56)
⇒ Cos A = -1/2
⇒ Cos A = Cos 2π/3
⇒ A = 2π/3
So, the greatest angle is = 2π/3
9. Answer: (b) tan 60
Explanation: tan20 × tan40 × tan80
= tan40 × tan80 × tan20
= [{sin40 × sin80}/{cos40 × cos80}] × (sin20/cos20)
= [{2 x sin40 × sin80}/{2 × cos40 × cos80}] × (sin20/cos20)
= [{cos40 – cos120}/{cos120 + cos40}] × (sin20/cos20)
= [{cos40 – cos (90 + 30)}/{cos(90 + 30) + cos40}] × (sin 20/cos 20)
= [{cos40 + sin30}/{-sin30 + cos40}] × (sin20/cos 20)
= [{(2 × cos40 + 1)/2}/{(-1 + cos40)/2}] × (sin20/cos20)
= [{2 × cos40 + 1}/{-1 + cos40}] × (sin 20/cos20)
= [{2 × cos40 × sin20 + sin20}/{-cos20 + cos40 × cos20}]
= (sin60 – sin20 + sin20)/(-cos20 + cos60 + cos20)
= sin60/cos60
= tan 60
10. Answer: (b) 1
Explanation: Given, cos 20 + 2sin2 55 – \(\sqrt2\;sin65\)
= cos20 + 1 – cos110 – \(\sqrt 2\;sin 65\) {since cos 2x = 1 – 2sin2 x}
= 1 + cos20 – cos 110 – √\(\sqrt \;sin 65\)
= 1 – 2 × sin {(20 + 110)/2 × sin{(20 – 110)/2} – \(\sqrt 2\;sin 65\) {Apply cos C – cos D formula}
= 1 – 2 × sin65 × sin (-45) – \(\sqrt2\;sin 65\)
= 1 + 2 × sin65 × sin45 – \(\sqrt 2\;sin 65\)
= 1 + (2 × sin 65)/\(\sqrt2\) – \(\sqrt 2\;sin 65\)
= 1 + \(\sqrt2\) ( sin 65 – √2 sin 65
= 1