TS Inter 1st Year Maths 1A Trigonometric Ratios upto Transformations Solutions Exercise 6(c)

I.
Question 1.
Simplify the following
(i) cos 100° . cos 40° + sin 100° . sin 40°

Answer:
Use cos A. cos B + sin A sin B = cos (A – B)
∴ cos 100° . cos 40° + sin 100°.sin 40°
= cos (100° – 40°)
= cos 60°
= $\frac{1}{2}$ = R.H.S

(ii) $\frac{\cot 55^{\circ} \cot 35^{\circ}-1}{\cot 55^{\circ}+\cot 35^{\circ}}$

Answer:

(iii) tan [$\frac{\pi}{4}$ + θ]. tan[$\frac{\pi}{4}$ – θ]
Answer:

(iv) Evaluate Σ$\frac{\cos ^2 \mathrm{~B}-\cos ^2 \mathrm{~A}}{\cos ^2 \mathrm{~A} \cos ^2 \mathrm{~B}}$ if none of sin A, sin B, sin C is zero.
Answer:
Σ$\frac{\cos ^2 \mathrm{~B}-\cos ^2 \mathrm{~A}}{\cos ^2 \mathrm{~A} \cos ^2 \mathrm{~B}}$ = Σ$\left(\frac{\sin C \cos A-\cos C \sin A}{\sin C \sin A}\right)$
= Σ (cot A – cot C)
= cot A – cot C + cot B – cot A + cot C – cot B = 0

Question 4.
(i) Prove that
cos 35° + cos 85° + cos 155° = 0

Answer:
cos 35° + cos 85° + cos 155°
= cos 35° + 2cos$\left(\frac{85+155}{2}\right)$cos$\left(\frac{85-155}{2}\right)$
= cos 35° + 2 cos 120° cos (-35°)
= cos 35° – cos 35° = 0

(ii) tan 72° = tan 18° + 2 tan 54°
Answer:
We have cot A – tan A = $\frac{1}{\tan A}$ – tan A
⇒ $\frac{1-\tan ^2 A}{\tan A}=\frac{2\left(1-\tan ^2 A\right)}{2 \tan A}$ = 2 cot 2A
∴ cot A – tan A = 2 cot 2A
⇒ cot A = tan A + 2 cot 2A°
Take A = 18°, then cot 18° = tan 18° + 2 cot 36°
⇒ cot(90 – 72) = tan 18° + 2 cot(90 – 54)
⇒ tan 72° = tan 18° + 2 tan 54°

(iii) sin 750° cos 480° + cos 120° cos 60° = –$\frac{1}{2}$
Answer:
L.H.S = sin [2.(360) + 30] cos[360 + 120] + cos 120 cos 60
= sin 30 cos 120 + cos 120 cos 60
= $\left(\frac{1}{2}\right)\left(-\frac{1}{2}\right)+\left(-\frac{1}{2}\right)\left(\frac{1}{2}\right)=-\frac{1}{2}$

(iv) cos A + cos($\frac{4 \pi}{3}$ – A) + cos($\frac{4 \pi}{3}$ + A) = 0
Answer:
Use cos(A + B) + cos(A – B) = 2cos A cos B
L.H.S = cos A + 2cos$\frac{4 \pi}{3}$cos A
= cos A + 2 cos 240 cos A
= cos A + 2 cos ( 180 + 60) cos A
= cos A + 2 ( – cos 60) cos A
= cos A + 2$\left(-\frac{1}{2}\right)$cos A = cos A – cos A = 0

(v) cos²θ + cos²($\frac{2 \pi}{3}$ + θ) + cos²($\frac{2 \pi}{3}$ – θ) = $\frac{3}{2}$
Answer:
cos²θ + cos²( 120 + θ) + cos²( 120 – θ)
= cos²θ + cos²(120 + θ) + 1 – sin²(120 – θ)
= 1 + cos²θ + cos [ 120 + θ + 120 – θ] cos [ 120 + θ- 120 + θ]
= 1 + cos²θ + cos (240) cos 2θ
= 1 + cos²θ + cos (180 + 60) cos 2θ
[∵ Use cos²A – sin²B = cos (A + B) cos (A – B)]
= 1 + cos²θ – cos 60 ( 2 cos2θ – 1)
= 1 + cos²θ – $\frac{1}{2}$ ( 2 cos2θ – 1)
= 1 + cos²θ – cos² θ + $\frac{1}{2}=\frac{3}{2}$

Question 5.
Evaluate
(i) sin²82$\frac{1}{2}^{\circ}$ – sin²22$\frac{1}{2}^{\circ}$

Answer:
sin²82$\frac{1}{2}^{\circ}$ – sin²22$\frac{1}{2}^{\circ}$
= sin[82$\frac{1}{2}^{\circ}$ + 22$\frac{1}{2}^{\circ}$]
∵ Use sin²A – sin²B = sin(A + B) sin(A – B)
= sin 105° . sin 60°
= sin 60° sin (60° + 45°)
= sin 60° [sin 60° cos 45° + cos 60° sin 45°]
= $\frac{\sqrt{3}}{2}\left[\frac{\sqrt{3}}{2 \sqrt{2}}+\frac{1}{2 \sqrt{2}}\right]=\frac{\sqrt{3}(\sqrt{3}+1)}{4 \sqrt{2}}=\frac{3+\sqrt{3}}{4 \sqrt{2}}$

(ii) cos² 112$\frac{1}{2}^{\circ}$ – sin²52$\frac{1}{2}^{\circ}$
Answer:
Use cos² A – sin² B = cos (A + B) cos (A – B)
cos²112$\frac{1}{2}^{\circ}$ – sin² 52$\frac{1}{2}^{\circ}$
= cos [112$\frac{1}{2}^{\circ}$ + 52$\frac{1}{2}^{\circ}$]cos[112$\frac{1}{2}^{\circ}$ -52$\frac{1}{2}^{\circ}$]
= cos 165° . cos 60
= cos 60° cos (180 – 15)
= -cos 60°. cos 15°
= $-\frac{1}{2}\left[\frac{\sqrt{3}+1}{2 \sqrt{2}}\right]=-\frac{(\sqrt{3}+1)}{4 \sqrt{2}}$

(iii) sin²$\left[\frac{\pi}{8}+\frac{A}{2}\right]$ – sin²$\left[\frac{\pi}{8}-\frac{A}{2}\right]$
Answer:

(iv) cos²52$\frac{1}{2}^{\circ}$ – sin²22$\frac{1}{2}^{\circ}$ (M’ 2010, 06 Jun 08)
Answer:
[∵ cos²A – sin²B = cos (A + B) cos (A – B)]
cos²52$\frac{1}{2}^{\circ}$ – sin²22$\frac{1}{2}^{\circ}$
= cos[52$\frac{1}{2}^{\circ}$ + 22$\frac{1}{2}^{\circ}$]cos[52$\frac{1}{2}^{\circ}$ – 22$\frac{1}{2}^{\circ}$]
= cos 75° cos 30°
= cos 30° cos(90 – 15)
= cos 30° sin 15°
= $\frac{\sqrt{3}}{2}\left(\frac{\sqrt{3}-1}{2 \sqrt{2}}\right)=\frac{3-\sqrt{3}}{4 \sqrt{2}}$

Question 6.
Find the minimum and maximum values of
(i) 3 cos x + 4 sin x

Answer:
Recall for a cos x + b sin x + c

(ii) sin 2x – cos 2x
Answer:
a = 1, b = -1, c = 0

Question 7.
Find the range of
(i) 7 cos x – 24 sin x + 5

Answer:
a = -24, b = 7, c = 5

(ii) 13 cos x + 3√3 sin x – 4
Answer:
a = 3√3, b = 13, c = -4

II.
Question 1.
(i)If cos α = –$\frac{3}{5}$ and sin β = $\frac{7}{25}$ where $\frac{\pi}{2}$ < α < π and 0 < β < $\frac{\pi}{2}$, then find the values of tan(α + β) and sin(α + β).

Answer:
cos α = –$\frac{3}{5}$, and α lies in second quadrant
sin α = $\frac{4}{5}$ ∴ tan α = –$\frac{4}{5}$

(ii) If 0 < A < B < $\frac{\pi}{4}$ and sin(A + B) = $\frac{24}{25}$ and cos (A – B) = $\frac{4}{5}$, then find the value of tan 2A. (March 2015-T.S)
Answer:

(iii) If A + B, A are acute angles such that sin (A + B) = $\frac{24}{25}$ and tan A = $\frac{3}{5}$, then find the value of cos B.
Answer:
A + B, A are acute angles ⇒ B is also acute.
Given sin (A + B) = $\frac{24}{25}$, we have

(iv) If tan α – tan β = m, and cot α – cot β = n then prove that cot(α – β) = $\frac{1}{\mathrm{~m}}-\frac{1}{\mathrm{n}}$.
Answer:
We have tan α – tan β = m

(v) If tan (α – β) = $\frac{7}{24}$ and tan α = $\frac{4}{3}$ where α and β are in the first quadrant prove that α + β = π/2.
Answer:

Question 2.
i) Find the expansion of sin (A + B – C).

Answer:
0 sin (A + B – C) = Sin [ (A + B) – C]
= sin (A + B) cos C – cos (A + B) sin C] – (sin A cos B + cos A sin B) cos C – (cos A cos B – sin A sin B] sin C
= sin A cos B cos C + cos A sin B cos C – cos A cos B sin C – sin A sin B sin C

ii) Find the expansion of cos (A – B – C).
Answer:
cos (A – B – C) = cos [(A – B) – C]
= cos (A – B) cos C + sin (A – B) sin C
= (cos A cos B + sin A sin B) cos C + (sin A cos B – cos A sin B] sin C
= cos A cos B cos C + sin A sin B cos C + sin A cos B sin C – cos A sin B sin C

iii) In a ΔABC, A is obtuse. If sin A = $\frac{3}{5}$ and sin B = $\frac{5}{13}$, then show that sin C = $\frac{16}{65}$
Answer:
Given, A + B + C = 180°
⇒ A + B = 180° – C
∴ sin (A + B) = sin (180° – C)
= sin C ………………..(1)
∴ A is obtuse angle and A lies in II quadrant.

iv) If $\frac{\sin (\alpha+\beta)}{\sin (\alpha-\beta)}=\frac{a+b}{a-b}$ then prove that a tan β = b tan α.
Answer: