Odisha State Board Elements of Mathematics Class 11 CHSE Odisha Solutions Chapter 4 Trigonometric Functions Ex 4(b) Textbook Exercise Questions and Answers.
CHSE Odisha Class 11 Math Solutions Chapter 4 Trigonometric Functions Exercise 4(b)
Question 1.
In the following questions, write ‘T’ for true and ‘F’ for false statements.
(i) If tan x + tan y = 5 and tan x, tan y = 1/2 then cot (x + y) = 10
Solution:
False
(ii) √3 (1 + tan 15°) = 1 – tan 15°
Solution:
False
(iii) If θ lies in 3rd quadrant, then cos \frac{\theta}{2} + sin \frac{\theta}{2} is positive.
Solution:
True
(iv) 2 sin 105°. sin 15° = 1/2.
Solution:
True
(v) If cos A = cos B = 1 then tan\frac{A+B}{2}. tan\frac{A+B}{2} = 1
Solution:
False
(vi) cos 15° cos7 \frac{1}{2}^{\circ}. sin 7 \frac{1}{2}^{\circ} = 1
Solution:
False
(vii) sin 20° (3 – 4 cos2 70°) = \frac{\sqrt{3}}{2}
Solution:
True
(viii) √3 (3 tan 10° – tan3 10°) = 1 – 3tan2 10°
Solution:
True
(ix) \frac{2 \tan 7 \frac{1^{\circ}}{2}\left(1-\tan ^2 7 \frac{1^{\circ}}{2}\right)}{\left(1+\tan ^2 7 \frac{1^{\circ}}{2}\right)^2} = 1
Solution:
False
(x) The minimum value of sin θ. cos θ is (-1)2.
Solution:
False
Question 2.
In the following questions, fill in the gaps with correct answers choice from the brackets.
(i) If α and β lie in 1st and 2nd quadrants respectively, and if sin α = 1/2, sin β = 1/3, then sin (α + β) = _______. \left(\frac{1}{2 \sqrt{3}}+\frac{\sqrt{2}}{3}, \frac{1}{2 \sqrt{3}}-\frac{\sqrt{2}}{3}, \frac{-1}{2 \sqrt{3}}+\frac{\sqrt{2}}{3}\right)
Solution:
\frac{1}{2 \sqrt{3}}+\frac{\sqrt{2}}{3}
(ii) if tan α = 1/2, tan β = 1/3, then α + β = ______ \left(\frac{\pi}{2}, \frac{\pi}{4}, \frac{\pi}{3}\right)
Solution:
\frac{\pi}{4}
(iii) The value of \frac{\cos 15^{\circ}+\sin 15^{\circ}}{\cos 15^{\circ}-\sin 15^{\circ}} = ______ \left(\frac{\sqrt{3}}{2}, \sqrt{3}, \frac{1}{\sqrt{3}}\right)
Solution:
√3
(iv) if \frac{1+\sin A}{\cos A} = √2 + 1, then the value of \frac{1-\sin A}{\cos A} is_________, \left(\frac{1}{\sqrt{2}-1}, \sqrt{2}-1, \sqrt{2}+1\right)
Solution:
√2 – 1
(v) sin 105°. cos 105° = \left(\frac{1}{2},-\frac{1}{4},-\frac{1}{2}\right)
Solution:
– 1/4
(vi) 2 sin67 \frac{1}{2}^{\circ} cos22 \frac{1}{2}^{\circ} = ___ \left(1-\frac{1}{\sqrt{3}}, 1+\frac{1}{\sqrt{2}},-1+\frac{1}{\sqrt{2}}\right)
Solution:
1 + \frac{1}{\sqrt{2}}
(vii) sin 35° + cos 5° =____ (2 cos 25°, √3 cos 25°, √3 sin 25°)
Solution:
√3 cos 25°
(viii) sin2 24° – sin2 26° =_____ \left(\frac{\sqrt{5}+1}{8}, \frac{\sqrt{5}-1}{8}, \frac{\sqrt{5}-1}{4}\right)
Solution:
\frac{\sqrt{5}-1}{8}
(ix) sin 70° (4 cos2 20° – 3) =_____ \left(\frac{\sqrt{3}}{2}, \frac{1}{2}, \sqrt{3}\right)
Solution:
1/2
(x) cos 3θ + sin 3θ is maximum if θ =_____ (60°, 15°, 45°)
Solution:
15°
(xi) sin 15° – cos 15° = _____ (1/2, 0, positive, negative)
Solution:
Negative
(xii) If θ lies in the third quadrant and tan θ = 2, then the value of sin θ is ____. \left(\frac{1}{\sqrt{5}},-\frac{1}{\sqrt{5}}, \frac{2}{\sqrt{5}}, \frac{-2}{\sqrt{5}}\right)
Solution:
\frac{-2}{\sqrt{5}}
(xiii) The correct expression is. (sin 1° > sin 1, sin 1° < sin 1, sin 1° = sin 1, sin 1° = \frac{\pi}{180^{\circ}} sin 1)
Solution:
sin 1° < sin 1
(xiv) The correct expression is —. (tan 1 > tan 2, tan 1 < tan 2, tan 1 = 1/2 tan 2, tan 1 < 0)
Solution:
tan 1 > tan 2
Question 3.
Prove the following
(i) sin A. sin (B – C) + sin B sin (C – A) + sin C. sin(A – B) = 0
Solution:
L. H. S
= sin A sin (B – C) + sin B sin (C – A) + sin C sin (A – B)
= sin A (sin B cos C – cos B sin C) + sin B (sin C cos A – cos C sin A) + sin C (sin A cos B – cos A sin B)
= sin A sin B cos C – sin A cos B sin C + cos A sin B sin C – sin A sin B cos C + sin A cos B sin C – cos A sin B sin C
= 0 = R. H. S
(ii) cos A. sin (B – C) + sin B sin (C – A) + cos C. sin(A – B) = 0
Solution:
L.H.S.
= cos A sin (B – C) + cos B sin (C – A) + cos C sin (A – B)
= cos A (sin B cos C – cos B sin C) + cos B (sin C cos A – cos C sin A) + cos C (sin A cos B – cos A sin B)
= cos A sin B cos C – cos A cos B sin C + cos A cos B sin C- sin A cos B cos C + sin A cos B cos C – cos A sin B cos C = 0 = R. H. S.
(iii) \frac{\sin (B-C)}{\sin B \cdot \sin C} + \frac{\sin (C-A)}{\sin C \cdot \sin A} + \frac{\sin (A-B)}{\sin A \cdot \sin B} = 0
Solution:
L. H. S.
(iv) tan2 A – tan2 B = \frac{\sin (\mathbf{A}+\mathbf{B}) \cdot \sin (\mathbf{A}-\mathbf{B})}{\cos ^2 \mathbf{A} \cdot \cos ^2 \mathbf{B}}
Solution:
= tan2 A sec2 B – tan2 B sec2 A
= tan2 A (1 + tan2 B) – tan2 B (1 + tan2 A)
= tan2 A + tan2 A tan2 B – tan2 B tan2A tan2 B
= tan2 A – tan2 B = L. H. S
Question 4.
Prove the following :
Solution:
(i) tan 75° + cot 75° = 4
Solution:
L.H.S = tan 75° + cot 75° = tan 75° + \frac{1}{\tan 75^{\circ}}
(ii) sin2 18° + cos2 36° = 3/4
Solution:
(iii) sin 18°. cos 36° = 1/4
Solution:
sin 18° cos 36°
=\left(\frac{\sqrt{5}-1}{4}\right)\left(\frac{\sqrt{5}+1}{4}\right)=\frac{5-1}{16}=\frac{1}{4}
(iv) sin 15° = \frac{\sqrt{3}-1}{2 \sqrt{2}}
Solution:
sin 15° = sin (45° – 30°)
= sin 45° cos 30° – cos 45° sin 30°
=\frac{1}{\sqrt{2}} \cdot \frac{\sqrt{3}}{2}-\frac{1}{\sqrt{2}} \cdot \frac{1}{2}=\frac{\sqrt{3}-1}{2 \sqrt{2}}
(v) cot \frac{\pi}{8} – tan \frac{\pi}{8} = 2
Solution:
(vi) \frac{\cos 9^{\circ}+\sin 9^{\circ}}{\cos 9^{\circ}-\sin 9^{\circ}} = tan 54°
Solution:
(vii) tan 10° + tan 35° + tan 10°. tan 35 = 1
Solution:
Question 5.
Prove the following:
(i) cot 2A = \frac{\cot ^2 A-1}{2 \cot A}
Solution:
(ii) \frac{\sin B}{\sin A}=\frac{\sin (2 A+B)}{\sin A} – 2cos(A + B)
Solution:
= \frac{\sin B}{\sin A} = L.H.S
(iii) \frac{\sin 2 A+\sin 2 B}{\sin 2 A-\sin 2 B}=\frac{\tan (A+B)}{\tan (A-B)}
Solution:
(iv) \frac{\cot A-\tan A}{\cot A+\tan A} = cos2A
Solution:
(v) \frac{\sin 2 A+\sin 5 A-\sin A}{\cos 2 A+\cos 5 A+\cos A} = tan 2A
Solution:
(vi) cot A – tan A = 2 cot 2A
Solution:
(vii) cot A – cosec 2A = cot 2A
Solution:
(viii) \frac{\cos A-\sin A}{\cos A+\sin A} = sec 2A – tan 2A
Solution:
(ix) tan θ (1 + sec 2θ) = tan 2θ
Solution:
(x) \frac{\sin A+\sin B}{\sin A-\sin B} = tan \frac{A+B}{2} . cot \frac{A-B}{2}
Solution:
(xi) sin 50° – sin 70° + sin 10° = 0
Solution:
L. H. S. = sin 50° – sin 70° + sin 10°
= sin (60° – 10°) – sin (60° + 10°) + sin 10°
= – 2 cos 60° sin 10° + sin 10°
= – 2 × 1/2 sin 10° + sin 10°
= – sin 10° + sin 10° = 0 = R. H. S.
(xii) cos 80° + cos 40° – cos 20° = 0
Solution:
L. H. S. = cos 80° + cos 40° – cos 20°
= cos(60° + 20°) + cos (60° – 20°) – cos 20°
= 2 cos 60° cos 20° – cos 20° = 0
(xiii) 8 sin 10°. sin 50°. sin 70° = 1
Solution:
L. H. S. = 8 sin 10° sin 50° sin 70°
= 8 sin 10° sin (60° – 10°) sin (60° + 10°)
= 8 sin 10° (sin2 60° – sin2 10°)
= 8 sin 10°(3/4 – sin2 10°)
= 6 sin 10° – 8 sin3 10°)
= 2 (3 sin 10° – 4 sin3 10°)
= 2 sin (3 × 10°)
= 2 sin 30° = 2 × 1/2 = 1 = R.H.S
(xiv) 4 sin A sin (60° – A) sin (60° + A) – sin 3A = 0
Solution:
L. H. S. = 4 sin A sin (60° – A)
sin (60° + A) – sin 3A
= 4 sin A (sin2 60°- sin2 A) – sin 3A
= 4 sin A. 3/4 – 4 sin3 A – sin 3A
= (3 sin A – 4 sin3A) – sin 3A
= sin 3A – sin 3A = 0
(xv) tan 3A – tan 2A – tan A = tan 3A tan 2A tan A
Solution:
We have tan 3A = tan (2A + A)
or, tan 3A = \frac{\tan 2 A+\tan A}{1-\tan 2 A \tan A}
or, tan 3A (1 – tan 2A tan A) = tan 2A + tan A
or, tan 3A – tan 3A tan 2A tan A = tan 2A + tan A
or, tan 3A – tan 2A – tan A = tan 3A tan 2A tan A (Proved)
Question 6.
Prove the following
(i) tan\frac{A}{2} = \sqrt{\frac{1-\cos A}{1+\cos A}}
Solution:
(ii) \sqrt{\frac{1+\sin A}{1-\sin A}} = tan\left(\frac{\pi}{4}+\frac{A}{2}\right)
Solution:
(iii) \frac{1+\tan \frac{A}{2}}{1-\tan \frac{A}{2}} = sec A + tan A
Solution:
(iv) sec θ + tan θ = tan\left(\frac{\pi}{4}+\frac{θ}{2}\right)
Solution:
(v) cot\frac{A}{2} = \frac{\sin A}{1-\cos A}
Solution:
R.H.S = \frac{\sin A}{1-\cos A}=\frac{2 \sin \frac{A}{2} \cos \frac{A}{2}}{2 \sin ^2 \frac{A}{2}}
= cot\frac{A}{2} R.H.S
Question 7.
Find the maximum value of the following.
(i) 5 sin x + 12 cos x
Solution:
5 sin x + 12 cos x
Let 5 = r cos θ, 12 = r sin θ 52
∴ 52 = r2 cos2 θ, 122 = r2 sin2 θ
∴ 52 + 122 = r2 (cos2 θ + sin2 θ) = r2
∴ r = \sqrt{25+144} = 13
∴ The maximum value of 5 sin x + 12 cos x is 13.
(ii) 24 sin x – 7 cos x
Solution:
The maximum value of 24 sin x – 7 cos x is
\sqrt{(24)^2+(-7)^2}
= \sqrt{576+49}=\sqrt{625} = 25
(iii) 2 + 3 sin x + 4 cos x
Solution:
The maximum value of 3 sin x + 4 cos x is
\sqrt{3^2+4^2} = 5
∴ Maximum value of 2 + 3 sin x + 4 cos x is 2 + 5 = 7
(iv) 8 cos x – 15 sin x – 2
Solution:
Maximum value of 8 cos x- 15 sin x is \sqrt{(8)^2+(-15)^2}
= \sqrt{64+225}=\sqrt{289} = 17
∴ Maximum value of 8 cos x- 15 sin x – 2 is 17 – 2 = 15
Question 8.
Answer the following:
(i) If tan A = \frac{13}{27}, tan B = \frac{7}{20} and A, B are acute, show that A + B = 45°.
Solution:
(ii) If tan θ = \frac{b}{a}, find the value of a cos 2θ + b sin 2θ.
Solution:
(iii) If sec A – tan A = \frac{1}{2} and 0< A < 90° then show that sec A = \frac{5}{4}
Solution:
If sec A – tan A = \frac{1}{2}, ….(1)
0< A < 90°
⇒ sec A – tan A = \frac{\sec ^2 A-\tan ^2 A}{2}
= \frac{(\sec \mathrm{A}+\tan \mathrm{A})(\sec \mathrm{A}-\tan \mathrm{A})}{2}
or, sec A + tan A = 2 ……(2)
Now adding eqn. (1) and (2), We have
2 sec A = \frac{1}{2} + 2 = \frac{5}{2}
or, sec A = \frac{5}{4}
(iv) If sin θ + sin Φ = a and cos θ + cos Φ = b the show that tan \frac{1}{2} (θ + Φ) \frac{a}{b}
Solution:
(v) If tan θ = \frac{a \sin x+b \sin y}{a \cos x+b \cos y} then show that a sin (θ – x) + b sin (θ – y) = 0
Solution:
tan θ = \frac{a \sin x+b \sin y}{a \cos x+b \cos y}
or, \frac{\sin \theta}{\cos \theta}=\frac{a \sin x+b \sin y}{a \cos x+b \cos y}
or, a sin θ cos x + b sin θ cos y = a cos θ sin x + b cos θ sin y
or, a (sin θ cos x – cos θ sin x) + b (sin θ cos y – cos θ sin y) = 0
or, a sin (θ – x) + b sin (θ – y) = 0
(vi) If A + C = B, show that tan A. tan B. tan C = tan B – tan A – tan C.
Solution:
A + C = B
or, tan (A + C) = tan B
or, \frac{\tan A+\tan C}{1-\tan A \tan C}
or, tan A + tan C = tan B – tan A tan B tan C
or, tan A tan B tan C = tan B – tan A – tan C
(vii) If tan A = \frac{1}{5}, tan B = \frac{2}{3} then show that cos 2A = sin 2B.
Solution:
tan A = \frac{1}{5}, tan B = \frac{2}{3}
∴ cos 2A = \frac{1-\tan ^2 \mathrm{~A}}{1+\tan ^2 \mathrm{~A}}
(viii) If cos 2A = tan2 B, then show that cos 2B = tan2 A In Δ ABC, prove that.
Solution:
(ix) tan\frac{B+C}{2} = cot\frac{A}{2}
Solution:
(x) cos (A + B) + sin C = sin (A + B) – cos C
If A + B + C = π and cos A = cos B. cos C show that (xi and xii)
Solution:
We have A + B = π – C or, cos (A + B)
or, cos (A + B)
= cos (π – C) = – cos C
and sin (A + B) = sin (π – C) = sin C
∴ cos (A + B) + sin C
= – cos C + sin (A + B)
= sin (A + B) – cos C.
(xi) tan B + tan C = tan A
Solution:
[∴ A + B + C = π ⇒ B + C = π – A
⇒ sin (B + C) = sin (π – A) = sin A]
L. H. S. = tan B + tan C
(xii) 2 cot B. cot C = 1
Solution:
We have A + B + C = π
⇒ B + C = π – A
⇒ cos (B + C) = cos (π – A) = – cosA
⇒ cos B cos C – sin B sin C = – cos B cos C
(∵ cos B cos C = cos A)
⇒ 2 cos B cos C = sin B sin C
⇒ \frac{2 \cos B \cos C}{\sin B \sin C} = 1
⇒ 2 cot B cot C = 1
Question 9.
Prove the following:
(i) cos (A – D) sin (B – C) + cos (B – D) sin (C – A) + cos (C- D) sin (A – B) = 0
Solution:
L. H. S. = cos (A – D) sin (B – C) + cos (B – D) sin (C – A) + cos (C – D) sin (A- B)
= \frac{1}{2} [2 cos (A- D) sin (B – C) + 2 cos (B – D) sin (C – A) + 2 cos (C – D) sin (A – B)]
= \frac{1}{2} [ sin (A- D + B – C) – sin (A – D – B + C) + sin (B – D + C – A)- sin (B- D- C + A) + sin ( C – D + A – B)- sin (C – D- A + B)]
= \frac{1}{2} × 0 = 0 = R.H.S
(ii) sin 2A + sin 2B + sin 2 (A – B) = 4 sin A. cos B. cos (A – B)
Solution:
L. H. S. = sin 2A + sin 2B + sin 2 (A – B)
= 2 sin \frac{2 A+2 B}{2} cos \frac{2 A-2 B}{2} + 2 sin (A – B) cos (A – B)
= 2 sin (A + B) cos (A- B) + 2 sin (A – B) cos (A – B)
= 2 cos (A – B)[ sin (A-+ B) + sin(A – B)]
= 2 cos (A – B) × 2 sin A cos B
= 4 sin A cos B cos (A – B) = R. H. S.
(iii) cos 2A + cos 2B + cos 2 (A – B) + 1 = 4 cos A. cos B. cos (A – B)
Solution:
cos 2A + cos 2B + cos 2 (A- B) + 1
= 2 cos \frac{2 A+2 B}{2} cos \frac{2 A-2 B}{2} + 2 cos2 (A – B)
= 2 cos (A + B) cos (A- B) + 2 cos2 (A – B)
= 2 cos (A – B) [cos (A + B) + cos (A – B)]
= 2 cos (A – B) x 2 cos A cos B
= 4 cos A cos B cos (A – B)
= R. H. S
(iv) sin 2A + sin 2B + sin 2C- sin 2 (A + B + C)
Solution:
L. H. S. = sin 2A + sin 2B + sin 2C- sin 2 (A + B + C)
= 2 sin \frac{2 A+2 B}{2} cos \frac{2 A-2 B}{2} = 2 cos \frac{2 \mathrm{C}+2(\mathrm{~A}+\mathrm{B}+\mathrm{C})}{2} × \frac{2 \mathrm{C}-2(\mathrm{~A}+\mathrm{B}+\mathrm{C})}{2}
= 2 sin (A + B) cos (A – B) + 2 cos (A + B + 2C) sin (A – B)
= 2 sin (A + B) [cos (A- B) – cos (A + B + 2C)]
= 2 sin (A + B) × 2 sin \frac{A-B+A+B+2 C}{2} sin \frac{A+B+2 C-A+B}{2}
= 4 sin (A + B) sin (C + A) sin (B + C) = R. H. S.
(v) sin A + sin 3A + sin 5A = sin 3A (1 + 2 cos 2A)
Solution:
L. H. S. = sin A + sin 3A + sin 5A
= sin 3A + sin 5A + sin A
= sin 3A + 2 sin \frac{5 \mathrm{~A}+\mathrm{A}}{2} cos \frac{5 \mathrm{~A}-\mathrm{A}}{2}
= sin 3A + 2 sin 3A cos 2A
= sin 3A (1 + 2 cos 2A) = R. H. S
(vi) sin A – sin 3A + sin 5A = sin 3A (2 cos 2A – 1)
Solution:
L. H. S. = sin A – sin 3A + sin 5A
= sin A + sin 5A – sin 3A
= sin 5A + sin A – sin 3A
= 2 sin \frac{5 \mathrm{~A}+\mathrm{A}}{2} cos \frac{5 \mathrm{~A}-\mathrm{A}}{2} – sin 3A
= 2 sin 3A cos 2A – sin 3A
= sin 3A (2 cos 2A – 1) = R. H. S.
(vii) cos (A + B) + sin (A – B) = 2 sin (45° + A) cos (45° + B)
Solution:
R. H. S. = 2 sin (45° + A) cos (45° + B)
= sin (45° + A + 45° + B) + sin (45° + A – 45° – B)
= sin (90° + A + B) sin (A – B)
= cos (A + B) sin (A – B) = L. H. S.
(viii) cos (120° + A) cos (120° – A) + cos (120° + A) cos A + cos A cos (120° – A) + \frac{3}{4} = 0
Solution:
L. H. S.
= cos (120° + A) cos (120° – A) + cos (120° + A) cos A + cos A cos (120° – A) + \frac{3}{4}
= cos2 A – sin2 120° + cos A [ cos (120° + A) + cos (120° – A)] + \frac{3}{4}
= cos2 A – \frac{3}{4} + cos A ( 2 cos 120°. cos A) + \frac{3}{4}
= cos2 A + 2 cos 120°. cos2 A
= cos2 A + 2 \left(-\frac{1}{2}\right) . cos2 A
= cos2 A – cos2 A = 0
(ix) cos 4A – cos 4B = 8 (cos A – cos B) (cos A + cos B) (cos A – sin B) (cos A + sin B)
Solution:
R. H. S. = 8 (cos A- cos B) (cos A + cos B)
(cos A – sin B) (cos A + sin B)
= 8 (cos2 A – cos2 B) (cos2 A – sin2 B)
= – 8(cos2 B – cos2 A)(cos2 A – sin2 B)
= – 8 sin (A +B) sin (A – B) cos (A + B) cos (A – B)
= – 2 x [ 2 sin (A + B) cos (A + B)] [2 sin (A – B) cos (A – B)]
= – 2 sin (2A + 2B) sin (2A – 2B)
= – [cos (2A + 2B – 2A + 2B) – cos (2A + 2B + 2A – 2B)]
= – cos 4B + cos 4A
= cos 4A – cos 4B = L. H. S.
Question 10.
Prove the following:
(i) \frac{1-\tan ^2\left(45^{\circ}-A\right)}{1+\tan ^2\left(45^{\circ}-A\right)} = sin 2A
Solution:
L.H.S \frac{1-\tan ^2\left(45^{\circ}-A\right)}{1+\tan ^2\left(45^{\circ}-A\right)} = sin 2A = cos 2(45° – A)
= cos (90° – 2A) = sin 2A = R. H. S.
(ii) \frac{\cos A+\sin A}{\cos A-\sin A}-\frac{\cos A-\sin A}{\cos A+\sin A} = 2 tan 2A
Solution:
(iii) \frac{1-\cos 2 A+\sin 2 A}{1+\cos 2 A+\sin 2 A} = tan A
Solution:
(iv) \frac{\sin (\mathbf{A}+\mathbf{B})+\cos (\mathbf{A}-\mathbf{B})}{\sin (\mathbf{A}-\mathbf{B})+\cos (\mathbf{A}+\mathbf{B})} = sec 2B + tan 2B
Solution:
(v) \frac{\cos 7 \alpha+\cos 3 \alpha-\cos 5 \alpha-\cos \alpha}{\sin 7 \alpha-\sin 3 \alpha-\sin 5 \alpha+\sin \alpha} = cot 2α
Solution:
L.H.S = \begin{array}{r}
\cos 7 \alpha+\cos 3 \alpha \\
-\cos 5 \alpha-\cos 3 \alpha \\
\hline \sin 7 \alpha-\sin 3 \alpha \\
-\sin 5 \alpha+\sin \alpha
\end{array}
(vi) \frac{\sin \theta+\sin 3 \theta+\sin 5 \theta+\sin 7 \theta}{\cos \theta+\cos 3 \theta+\cos 5 \theta+\cos 7 \theta} = tan 4θ
Solution:
Question 11.
Prove the following:
(i) Express 4 cos A. cos B. cos C as the sum of four cosines.
Solution:
4 cos A cos B cos C
= 2 (2cos A cos B) cos C
= 2 [ cos (A + B) + cos (A – B)] cos C
= 2 cos (A + B) cos C + 2 cos (A – B) cos C
= cos (A + B + C) + cos (A + B – C) + cos (A – B + C) + cos (A – B – C)
(ii) Express cos 2A + cos 2B + cos 2C + cos 2 (A + B + C) as the product of three cosines.
Solution:
cos 2A + cos 2B + cos 2C + cos 2(A + B + C)
= 2 cos \frac{2 \mathrm{~A}+2 \mathrm{~B}}{2} cos \frac{2 \mathrm{~A}-2 \mathrm{~B}}{2} = 2 cos \frac{2 C+2(A+B+C)}{2} × cos \frac{2 C-2(A+B+C)}{2}
= 2 cos (A + B) cos (A- B) + 2 cos (A + B + 2C) cos (A + B)
= 2 cos (A + B) [cos (A- B) + cos (A + B + 2C)]
= 2 cos (A + B) × 2 cos \frac{(A-B+A+B+2 C)}{2} cos \frac{(A-B-A-B-2 C)}{2}
= 4 cos (A + B) cos (C + A) cos (B + C)
Question 12.
Prove the following:
(i) cos6 A – sin6 A = cos 2A(1 – \frac{1}{4} sin2 2A)
Solution:
L.H.S = cos6 A – sin6 A
= (cos2 A)3– (sin2 A)3 = (cos2 A – sin2 A)3 + 3 cos2 A sin2 A (cos2 A – sin2 A)
= cos3 2A + \frac{3}{4} sin2 2A cos 2A
= cos2 A (cos2 2A + \frac{3}{4} sin2 2A)
= cos 2A (1 – sin2 2A + \frac{3}{4} sin2 2A)
= cos 2A (1 – \frac{1}{4} sin2 2A) = R.H.S
(ii) cos6A + sin6 A = \frac{1}{4} (1 + 3 cos22A)
Solution:
L.H.S = cos6 A + sin6 A
= (cos2 A)3 + (sin2 A)3
= (cos2 A)3 + (sin2 A)3 – 3 cos2 A. sin2 A (cos2 A + sin2 A)
= 1 – \frac{3}{4} sin2 2A = 1 – \frac{3}{4} (1 – cos2 2A)
= 1 – \frac{3}{4} + \frac{3}{4} cos2 2A
= \frac{1}{4} + \frac{3}{4} cos2 2A = \frac{1}{4} (1 + 3cos2 2A) = R.H.S.
(iii) cos3 A. cos 3A + sin3 A sin3 A = cos3 2A
Solution:
L.H.S = cos3 A cos 3A + sin3 A sin 3A
= cos3 A (4 cos3 A- 3 cos A) + sin3 A (3 sin A- 4 sin3 A)
= 4 cos6A- 3 cos4A + 3 sin4A- 4 sin6A
= 4 (cos6A- sin6A) – 3(cos4A- sin4A)
= 4 {(cos2A)3– (sin2A)3} – 3 {(cos2A)2– (sin2A)2}
= 4 (cos2A- sin2A) {(cos2A)2 + cos2A sin2A + (sin2A)2} – 3 (cos2A- sin2A) (cos2A + sin2A)
= (cos2A- sin2A) [4 {(cos2A + sin2A)2-2 cos2A sin2A + cos2A sin2A} -3×1]
= cos 2A ( 4- 4 sin2A cos2A- 3)
= cos 2A ( 1 – 4 sin2A cos2A)
= cos 2A (1 – sin22A)
= cos 2A cos22A = cos32A = R. H. S.
(iv) sin4 θ = \frac{3}{8} – \frac{1}{2} cos 2θ + \frac{1}{8} cos 4θ
Solution:
R.H.S = \frac{3}{8} – \frac{1}{2} cos 2θ + \frac{1}{8} cos 4θ
= \frac{3}{8} – \frac{1}{2} (1 – 2 sin2 θ) + \frac{1}{8} (1 – 2 sin2 θ)
= \frac{3}{8} – \frac{1}{2} + sin2 θ + \frac{1}{8} – \frac{1}{4} sin2 2θ
= sin2 θ – \frac{1}{4} × 4 sin2 θ cos2 θ
= sin2 θ (1 – cos2 θ)
= sin2 θ sin2 θ = sin4 = L.H.S
(v) cot 3A = \frac{\cot ^3 A-3 \cot A}{3 \cot ^2 A-1}
Solution:
(vi) tan 4θ = \frac{4 \tan \theta-4 \tan ^3 \theta}{1-6 \tan ^2 \theta+\tan ^4 \theta}
Solution:
(vii) \frac{1}{\tan 3 A-\tan A}-\frac{1}{\cot 3 A-\cot A} = cot 2A
Solution:
(viii) \frac{\cot A}{\cot A-\cot 3 A}-\frac{\tan A}{\tan 3 A-\tan A} = 1
Solution:
Question 13.
Find the value of
sin 3°,cos 3°, 2 sin \frac{\pi}{32}
Solution:
sin 3° = sin (18° – 15°)
= sin 18° cos 15° – cos 18° sin 15°
Question 14.
If sin A + sin B = a and cos A + cos B = b, then show that
(i) tan(A + B) = \frac{2 a b}{b^2-a^2}
Solution:
(ii) sin (A + B) = \frac{2 a b}{b^2-a^2}
(iii) cos (A + B) = \frac{b^2-a^2}{b^2+a^2}
Question 15.
Prove the following:
(i) \frac{1+\sin A-\cos A}{1+\sin A+\cos A} = tan \frac{A}{2}
Solution:
(ii) 8 \sin ^4 \frac{1}{2} \theta-8 \sin ^2 \frac{1}{2} \theta + 1 = cos 2θ
Solution:
(iii) \cos ^4 \frac{\pi}{8}+\cos ^4 \frac{3 \pi}{8}+\cos ^4 \frac{5 \pi}{8} +\cos ^4 \frac{7 \pi}{8}=\frac{3}{2}
Solution:
(iv) cos2 \frac{\alpha}{2} (1- 2cos α)2 + sin2 α(1+ 2 cos α)2 =1
Solution:
Question 16.
Prove the following:
(i) sin 20°. sin 40°. sin 60°. sin 80° = \frac{3}{16}
Solution:
(ii) cos 36°. cos 72°. cos 108°. cos144° = \frac{7}{16}
Solution:
(iii) cos 10°. cos 30°. cos 50°. cos 70° = \frac{3}{16}
Solution:
(iv) cos 20°. cos 40°. cos 60°. cos 80° = \frac{1}{16}
Solution:
(v) tan 6°. tan 42°.tan 66°. tan 78° = 1 [Hints: Use the identity tan 3A = tan A tan (60° – A) tan (60° + A)]
Solution:
We have tan 3A
= tan A tan (60° – A) tan (60° + A)
Now putting A = 6° and 18°.
in (1) we have
tan 18° = tan 6° tan 54° tan 66° …(2)
and tan 54°
= tan 18° tan 42° tan 78° ……(3)
Multiplying (2) and (3) we have
tan 18° tan 54°
= tan 6° tan 54° tan 66° tan 18° . tan 42°. tan 78°
or, 1 = tan 6° tan 42° tan 66° tan 78°.
Question 17.
Prove the following:
(i) cos 7 \frac{1}{2}° = √6 + √3 + √2 + 2
Solution:
(ii) cot 22 \frac{1}{2}° = √2 + 1
Solution:
(iii) cot 37 \frac{1}{2}° = √6 – √3 – √2 + 2
Solution:
(iv) tan 37 \frac{1}{2}° = √6 + √3 – √2 + 2
Solution:
(v) cos \frac{\pi}{16} = 1/2 \sqrt{2+\sqrt{2+\sqrt{2}}}
∴ 2 cos \frac{\pi}{16}
Question 18.
(i) If sin A = K sin B, prove that tan 1/2(A – B) = \frac{K-1}{K+1} tan 1/2(A – B)
Solution:
(ii) Ifa cos (x + α) = b cos (x – α) show that (a + b) tan x = (a – b) cot α.
Solution:
or, cot x cot α = \frac{a+b}{a-b}
or, (a – b) cot α = \frac{a+b}{\cot x} = (a + b) tan x
or, (a + b) tan x =(a-b) cot α
(iii) An angle 0 is divided into two parts α, β such that tan α: tan β = x : y.
Prove that sin (α – β) = \frac{x-y}{x+y} sin θ.
Solution:
(iv) If sin θ + sin Φ = a, cos θ + cos Φ = b, show that
\frac{\sin \frac{\theta+\phi}{2}}{a}=\frac{\cos \frac{\theta+\phi}{2}}{b}=2 \frac{\cos \frac{\theta-\phi}{2}}{a^2-b^2}
Solution:
sin θ + sin Φ = a
cos θ + cos Φ = b
We have
a2 + b2 = (sin θ + sin Φ)2 + (sin θ + sin Φ)2
= sin2 θ + sin2 Φ + 2 sin θ. sin Φ + cos2 θ +cos2 Φ + 2 cos θ. cos Φ
= 2 + 2 (cos θ cos Φ + sin θ sin Φ)
= 2 + 2 cos (θ – Φ)
= 2 [1+ cos (θ – Φ)]
(v) If a cos α + b sin α = c = a cos β + b sin α then prove that
\frac{a}{\cos \frac{1}{2}(\alpha+\beta)}=\frac{b}{\sin \frac{1}{2}(\alpha+\beta)} =\frac{c}{\cos \frac{1}{2}(\alpha-\beta)}
Solution:
(vi) Prove that \left(\frac{\cos A+\cos B}{\sin A-\sin B}\right)^n+\left(\frac{\sin A+\sin B}{\cos A-\cos B}\right)^n = 2 cosn \frac{A-B}{2} or zero according as n is even or odd.
Solution:
Question 19.
(i) If (1 – e) tan2 \frac{\boldsymbol{\beta}}{2} = (1 + e) tan2 \frac{\boldsymbol{\alpha}}{2}, prove that cos β = \frac{\cos \alpha-e}{1-e \cos \alpha}
Solution:
(ii) If cos θ = \frac{\cos \mathbf{A}-\cos \mathbf{B}}{1-\cos \mathbf{A} \cdot \cos \mathbf{B}} prove that one of the values of
tan \frac{θ}{2} is tan \frac{A}{2} . tan \frac{B}{2}.
Solution:
(iii) If tan θ = \frac{\sin x \cdot \sin y}{\cos x+\cos y} then prove that one of the values of tan \frac{1}{2} θ tan \frac{1}{2} x and tan \frac{1}{2} y.
Solution:
(iv) If sec (Φ + a) + sec (Φ – α) = 2 sec Φ. show that cos Φ = ± √2 cos \frac{α}{2}
Solution:
(v) If tan A + tan B = a and cot A + cot B = b. then show that cot (A + B) = \frac{1}{a} – \frac{1}{b}.
Solution:
(vi) If cot θ = cos (x + y) and cot Φ = cos (x – y) show that tan (θ = Φ) = \frac{2 \sin x \cdot \sin y}{\cos ^2 x+\cos ^2 y}
Solution:
(vii) If tan β = \frac{n^2 \sin \alpha \cdot \cos \alpha}{1-n^2 \sin ^2 \alpha}, then show that \frac{\tan (\alpha-\beta)}{\tan \alpha} = 1 – n2
Solution:
(viii)If 2 tan α = 3 tan β, then prove that tan(α – β) = \frac{\sin 2 \beta}{5-\cos 2 \beta}
Solution:
(ix) If α, β are acute angles and cos 2α = \frac{3 \cos 2 \beta-1}{3-\cos 2 \beta} then prove that tan α = √2 tan β.
Solution:
Question 20.
If A + B + C = π, then prove the following.
(i) cos 2A + cos 2B + cos 2C + 1 + 4 cos A. cos B. cos C = 0
Solution:
A + B + C = π or, A + B = π – C
or, cos (A + B) = cos (π – C) = – cos C
∴ cos 2A + cos 2B + cos 2C
= 2 cos \frac{2 \mathrm{~A}+2 \mathrm{~B}}{2} cos \frac{2 \mathrm{~A}-2 \mathrm{~B}}{2}
= 2 cos (A + B) cos (A- B) + 2 cos2 C – 1
= – 2 cos C cos (A – B) + 2 cos2 C – 1
= – 1 – 2 cos C cos (A – B) – cos C
= – 1 – 2 cos C [cos (A- B) + cos (A + B)]
= – 1 – 2 cos C × 2 cos A cos B
= – 1 – 4 cos A cos B cos C
∴ L. H. S. = cos 2A + cos 2B + cos 2C + 1 + 4 cos A cos B cos C
= – 1 – 4 cos A cos B cos C + 1 + 4 cos A cos B cos C = 0 = R. H. S.
(ii) sin 2A + sin 2B – sin 2C = 4 cos A. cos B. sin C
Solution:
L. H. S. = sin 2A + sin 2B – sin 2C
= 2 sin \frac{2 A+2 B}{2} cos \frac{2 A-2 B}{2} – sin 2C
= 2 sin(A + B) cos (A – B) -2 sinC cos C
= 2 sin C cos (A- B) – 2 sin C cos C [ ∵ A + B= π – C]
or, sin (A + B) = sin (n- C) = sin C]
= 2 sin C [cos (A – B)- cos C]
= 2 sin C [cos (A – B) – cos (A + B)]
= 2 sin C. 2 cos A. cos B
= 4 cos A cos B sin C = R. H. S.
(iii) cos A + cos B + cos C = 1 + 4 sin \frac{1}{2} A. sin \frac{1}{2} B. sin \frac{1}{2} C.
Solution:
(iv) sin A + cos B- sin C = 4 sin \frac{1}{2} A. sin \frac{1}{2} B. cos \frac{1}{2} C.
Solution:
(v) cos2 A + cos2 B + 2cos A. cos B. cos C = sin2 C
Solution:
(vi) sin2 \frac{A}{2} + sin2 \frac{B}{2} + sin2 \frac{C}{2} = 1 – 2 sin \frac{A}{2}. sin \frac{B}{2}. sin \frac{C}{2}
Solution:
(vii) sin \frac{A}{2} + sin \frac{B}{2} + sin \frac{c}{2} = 4 sin \frac{\pi-A}{4} sin \frac{\pi-B}{4}. sin \frac{\pi-C}{4} + 1
Solution:
(viii) cos2 \frac{A}{2} + cos2 \frac{B}{2} – cos2 \frac{C}{2} = 2 cos \frac{A}{2}. cos \frac{B}{2}. sin \frac{C}{2}
Solution:
(ix) sin (B + 2C) + sin (C + 2A) + sin (A + 2B) = 4 sin \frac{B-C}{2}. sin \frac{C-A}{2}. sin \frac{A-B}{2}
Solution:
Question 21.
(i) Show that (2 cos θ – 1) (2 cos 2θ – 1) (2 cos2 2θ – 1) ….. (2 cos 2n-1 θ – 1) =\frac{2 \cos 2^n \theta+1}{2 \cos \theta+1}
Solution:
We have
(2 cos θ + 1) (2 cos θ – 1)
= 4 cos2 θ – 1=4 cos2 θ – 2+1
= 1 + 2(2 cos2 θ – 1)
= 1+2 cos 2θ = 2 cos 2θ+1
And, (2 cos 2θ + 1) (2 cos 2θ – 1)
= 4 cos2 2θ – 1 = 2 ( 2 cos2 2θ – 1) + 1
= 2 cos 4 θ + 1 = 2 cos 22 θ + 1
Similarly, we can prove,
( 2 cos 22 θ + 1) ( 2 cos 22 θ- 1)
= 2 cos 23 θ + 1
Proceeding in this way, we can prove,
(2 cos 2θ + 1) (2 cos 2θ – 1)
(2 cos 2θ- 1) (2 cos 22 θ – 1) …… (2 cos 2n-1 θ-1)
= 2 cos 2n θ + 1
or, (2 cos θ-1) (2 cos 2θ-1) (2 cos 22 θ- 1) (2 cos 2n-1 θ-1)
=\frac{2 \cos 2^n \theta+1}{2 \cos \theta+1}
(ii) Show that 2n cos θ. cos 2θ. cos22 θ ……… 2n-1 θ = 1 If θ = \frac{\pi}{2^n+1}
Solution:
we have, θ = \frac{\pi}{2^n+1}
or, 2n θ + θ = π
or, 2n θ = π – θ
or, 2n θ = sin (π – θ) = sin θ
or, \frac{\sin 2^n \theta}{\sin \theta} = 1 …….(1)
Again, sin 2n θ = 2 sin 2n-1 θ cos 2n-1 θ
= 2 × 2 sin 2n-2 θ cos 2n-2 θ cos 2n-1 θ
= 22 × 2 sin 2n-3 θ cos 2n-3 θ cos 2n-2 θ cos 2n-v θ
= 23 sin 2n-3 θ cos 2n-3 θ cos 2n-2 θ cos 2n-1 θ
………………………..
………………………..
………………………..
= 2n sin θ cos θ cos 2θ cos 22 θ …….. cos 2n-2 θ cos 2n-1 θ
or, 2n cos θ cos 2θ cos 22 θ …… cos 2n-1 θ = 1
(iii) Prove that \frac{\tan 2^n \theta}{\tan \theta} = = (1 + sec 2θ) (1 + sec22 θ) …. (1 + sec2n θ)
Solution:
Question 22.
If x + y + z = xyz, prove that
(i) \frac{x}{1-x^2}+\frac{y}{1-y^2}+\frac{z}{1-z^2} = =\frac{4 x y z}{\left(1-x^2\right)\left(1-y^2\right)\left(1-z^2\right)}
Solution:
x + y + z = xyz (Given)
Let x = tan α, y = tan β, z = tan γ
∴ tan α + tan β + tan γ = tan α .tan β .tan γ or; tan α + tan β
= – tan γ + tan α tan β tan γ
= – tan γ(1 – tan α tan β)
or, \frac{\tan \alpha+\tan \beta}{1-\tan \alpha \tan \beta} = – tan γ
or, tan (α + β) = tan (π- γ)
or, α + β = π-γ
or, 2α + 2β = 2π- 2γ
or, tan (2α + 2β = tan (2π- 2γ)
or, tan (2α + 2β) = tan (2π- 2γ)
(ii) \frac{3 x-x^3}{1-3 x^2}+\frac{3 y-y^3}{1-3 y^2}+\frac{3 z-z^3}{1-3 z^2} = \frac{3 x-x^3}{1-3 x^2} \cdot \frac{3 y-y^3}{1-3 y^2} \cdot \frac{3 z-z^3}{1-3 z^2}
Solution:
Question 23.
If \text { If } \frac{\sin ^4 \alpha}{a}+\frac{\cos ^4 \alpha}{b}=\frac{1}{a+b} show that \frac{\sin ^8 \alpha}{a^3}+\frac{\cos ^8 \alpha}{b^3}=\frac{1}{(a+b)^3}
Solution: