TS Inter 1st Year Maths 1A Ratios up to Transformations Important Questions Very Short Answer Type

Students must practice these Maths 1A Important Questions TS Inter 1st Year Maths 1A Ratios up to Transformations Important Questions Very Short Answer Type to help strengthen their preparations for exams.

TS Inter 1st Year Maths 1A Ratios up to Transformations Important Questions Very Short Answer Type

Question 1.
If cos θ = t(0 < t < 1) and θ does not lies in the first quadrant, find the values of sin θ and tan θ. [Mar. ’17(AP)]
Answer:
cos θ = t, (0< t < 1)
⇒ cos θ is positive and θ does not lie in first quadrant
⇒ θ lies in IVth quadrant
(a) sin θ = \(-\sqrt{1-\cos ^2 \theta}=-\sqrt{1-t^2}\)
(b) tan θ = \(-\sqrt{1-\cos ^2 \theta}=-\sqrt{1-t^2}\)

Question 2.
Find the value of sin²\(\frac{\pi}{10}\) + sin²\(\frac{4 \pi}{10}\) + sin²\(\frac{6 \pi}{10}\) + sin²\(\frac{9 \pi}{10}\)
Answer:

Question 3.
If sin θ = –\(\frac{1}{3}\) and θ does not lie in the third quadrant, find the value of cos θ, cot θ. [Mar. ’19(TS); Mar. ’13]
Answer:
sin θ = –\(\frac{1}{3}\) and sin θ is negative and does not lie in third quadrant, ⇒ θ lies in fourth quadrant. In IV quadrant cos θ is positive.
cos θ = \(\sqrt{1-\sin ^2 \theta}=\sqrt{1-\frac{1}{9}}=\frac{2 \sqrt{2}}{3}\)
cot θ = \(\sqrt{1-\sin ^2 \theta}=\sqrt{1-\frac{1}{9}}=\frac{2 \sqrt{2}}{3}\)

If sin θ = \(\frac{4}{5}\) and θ Is not In the first quadrant, find the value of cos θ. [Mar. 19 (AP): Mar. ’17 (TS)]
Answer:
\(\frac{-3}{5}\)

Question 4.
If sec θ + tan θ = \(\frac{2}{3}\), find the value of sin θ and determine the quadrant in which θ lies.
Answer:
sec θ + tan θ = \(\frac{2}{3}\) and sec²θ – tan²θ = 1
⇒ sec θ – tan θ = \(\frac{3}{2}\)
∴ 2sec θ = \(\frac{13}{6}\) ⇒ sec θ = \(\frac{13}{12}\)

Also 2 tan θ = \(\frac{2}{3}-\frac{3}{2}=-\frac{5}{6}\)
tan θ = \(-\frac{5}{6}\)

Now sin θ = \(\frac{\tan \theta}{\sec \theta}=\frac{-(5 / 12)}{13 / 12}=-\frac{5}{13}\)
Since tan θ is negative, sec is positive
∴ θ lies in fourth quadrant.

If cosec θ + cot θ = \(\frac{1}{3}\), find cos θ and determine the quadrant in which θ lies.
Answer:
\(\frac{-4}{5}\), Q₂

If sec θ + tan θ = 5, find the quadrant in which θ lies and find the value of sin θ. [May ’00]
Answer:
\(\frac{12}{13}\), Q₁

Question 5.
Show that cot\(\left(\frac{\pi}{20}\right)\).cot\(\left(\frac{3 \pi}{20}\right)\).cot\(\left(\frac{5 \pi}{20}\right)\).cot\(\left(\frac{7 \pi}{20}\right)\).cot\(\left(\frac{9 \pi}{20}\right)\) = 1. [Mar. ’05; May ’98]
Answer:
cot\(\left(\frac{\pi}{20}\right)\).cot\(\left(\frac{3 \pi}{20}\right)\).cot\(\left(\frac{5 \pi}{20}\right)\).cot\(\left(\frac{7 \pi}{20}\right)\).cot\(\left(\frac{9 \pi}{20}\right)\)
= cot 9°. cot 27°. cot 45°. cot 63°. cot 81°
= cot 9°. cot 27°. 1.cot (90 – 27) . cot (90 – 9)
= cot 9°. cot 27°. 1. tan 27°. tan 9° = 1

Question 6.
If 3 sin θ + 4 cos θ = 5, then find the value of 4sin θ – 3 cos θ. [Mar. ’12]
Answer:
Given 3 sin θ + 4 cos θ = 5 and suppose 4 sin θ – 3 cos θ = x
Squaring on adding
∴ (3sin θ + 4cosθ)² + (4sin θ – 3cosθ)² = 25 + x²
⇒ 9sin²θ + 24 sin θcos θ + 16 cos²θ + 16sin²θ – 24sinθcosθ + 9cos²θ = 25 + x²
⇒ 25 (sin²θ + cos²θ) = 25 + x
⇒ x = 0 ⇒ x = 0
∴ 4sin θ – 3cosθ = 0

Question 7.
If cos θ + sin θ = √2 cos θ , then show that cos θ – sin θ = √2 sin θ. [Mar ‘ 15(TS); May ’11; Mar. ’09, ’08]
Answer:
Given cos θ + sin θ = √2 cos θ
⇒ √2 cos θ – cos θ = sin θ
⇒ (√2 – 1)cos θ = sin θ
⇒ cos θ = \(\frac{\sin \theta}{\sqrt{2}-1}=\frac{\sqrt{2}+1}{(2-1)}\)sin θ
= (√2 + 1)sin θ = √2sin θ + sin θ
⇒ cos θ – sin θ = √2 sin θ

Question 8.
If tan 20° = λ tehn show that \(\frac{\tan 160^{\circ}-\tan 110^{\circ}}{1+\tan 160^{\circ} \tan 110^{\circ}}=\frac{1-\lambda^2}{2 \lambda}\). [Mar. ’05, Mar. ’16(AP)]
Answer:

Question 9.
Find the value of sin 330°. cos 120° + cos 210°. sin 300°. [Mar. ’18(AP)]
Answer:
sin 330° cos 120° + cos 210° sin 300°
= sin (360 – 30) cos (180 – 60) + cos (180 + 30) sin (360 – 60)
= (-sin 30°) (-cos 600) + (-cos 30°) (-sin 600)
= sin 30° cos 60° + cos 30° sin 60°
= sin (30 + 60)
= sin 90° = 1

Question 10.
If sin α + cosec α = 2, find the value of sinⁿα + cosecⁿα; n ∈ Z.
Answer:
Given sin α + cosec α = 2
Squaring on both sides sin² α + cosec² α = 2
=4 ⇒ sin²α + cosec²α = 2

Cubing of both sides
sin³ α + cosec³ α + 3 sin α cosec α (sin α cosec α) = 8
sin³α cosec³α + 3(2) = 8
⇒ sin³α + cosec³α = 2
In the same way sinⁿα + cosec³α = 2(n ∈ Z)

Question 11.
W Prove that sin 780°. sin 480° + cos 240°. cos 300° = \(\frac{1}{2}\)
Answer:
LHS = sin 780°. sin 480° + cos 240°. cos 300°
sin [2 × 360 + 60] sin [360 + 120] + cos [180 + 60] cos [360 – 60]
= sin 60 sin 120 – cos 60 cos 60
= sin 60 sin 60 – cos 60. cos 60
= \(\frac{\sqrt{3}}{2} \cdot \frac{\sqrt{3}}{2}-\frac{1}{2} \cdot \frac{1}{2}=\frac{3}{4}-\frac{1}{4}=\frac{1}{2}\)

Question 12.
Simplify: \(\frac{\sin \left(-\frac{11 \pi}{3}\right) \tan \left(\frac{35 \pi}{6}\right) \sec \left(-\frac{7 \pi}{3}\right)}{\cot \left(\frac{5 \pi}{4}\right) {cosec}\left(\frac{7 \pi}{4}\right) \cos \left(\frac{17 \pi}{6}\right)}\)
Answer:
sin\(\left(-\frac{11 \pi}{3}\right)\) = sin(-660)
= sin(-2 × 360° + 60°) = sin 60° = \(\frac{\sqrt{3}}{2}\)

tan \(\left(\frac{35 \pi}{6}\right)\) = tan(105)
= tan(3 × 360° – 30°) = -tan 30° = \(\frac{\sqrt{3}}{2}\)

sec\(\left(-\frac{7 \pi}{3}\right)\) = sec(-420°)
= sec 420° = sec(360 + 60) = sec 60° = 2

cot\(\left(\frac{5 \pi}{4}\right)\) = cot(225°)
= (cot (180 + 45) = cot 45° = 1

cosec\(\left(\frac{7 \pi}{4}\right)\) = cosec (315°)
= cosec(270 + 45) = -sec 45° = √2

cos\(\left(\frac{17 \pi}{6}\right)\) = cos(570)
= cosec(540 + 30) = -cos 30 = –\(\frac{\sqrt{3}}{2}\)

Question 14.
If A, B, C, D are angles of a cyclic quadrilateral, then prove that cos A + cos B + cos C + cos D = 0.
Answer:
A, B, C, D are angles of a cyclic quadrilateral
⇒ A + C = 180° and B + D = 180° …………..(1)
C = 180 – A and D = 180° – B
LHS = cos A + cos B + cos C + cos D
= cos A + cos B + cos (180 – A) + cos (180 – B)
= cos A + cos B – cos A – cos B = 0 = RHS

Question 15.
Prove that cos⁴α + 2 cos²α(1 – \(\frac{1}{\sec ^2 \alpha}\)) = 1 – sin⁴α.
Answer:
LHS = cos⁴α + 2 cos²α(1 – \(\frac{1}{\sec ^2 \alpha}\))
= cos⁴α + 2cos²α(1 – cos²α)
= cos⁴α + 2 cos²α sin²α
=cos²α [cos²α + 2sin²α]
= (1 – sin2 a) [cos²α + sin²α + sin²α]
= (1 – sin²α)(1 + sin²α)= 1 – sin⁴α

Question 16.
If \(\frac{2 \sin \theta}{(1+\cos \theta+\sin \theta)}\) = x, then find the value of \(\frac{1-\cos \theta+\sin \theta}{1+\sin \theta}\)
Answer:

Question 17.
Prove that sin²52\(\frac{1}{2}^{\circ}\) + sin²22\(\frac{1}{2}^{\circ}=\frac{\sqrt{3}+1}{4 \sqrt{2}}\)
Answer:

Evaluate sin²82\(\frac{1}{2}^{\circ}\) – sin²22\(\frac{1}{2}^{\circ}\)
Answer:
\(\frac{\sqrt{3}(\sqrt{3}+1)}{4 \sqrt{2}}\)

Question 18.
Evaluate cos²52\(\frac{1}{2}^{\circ}\) – sin²22\(\frac{1}{2}^{\circ}\)
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°
= \(\frac{\sqrt{3}}{2}\left(\frac{\sqrt{3}-1}{2 \sqrt{2}}\right)=\frac{3-\sqrt{3}}{4 \sqrt{2}}\)

Evaluate cos²112\(\frac{1}{2}^{\circ}\) – sin²52\(\frac{1}{2}^{\circ}\).
Answer:
\(-\frac{\sqrt{3}+1}{4 \sqrt{2}}\)

Question 19.
Prove that tan 70° – tan 20° = 2tan 50°
Answer:

Question 20.
What is the value of tan 20° + tan 40° + √3 tan 20° tan 40°
Answer:
We have 20° + 40° = 60°
∴ tan(20° + 40°) = tan 60°
⇒ \(\frac{\tan 20^{\circ}+\tan 40^{\circ}}{1-\tan 20^{\circ} \tan 40^{\circ}}\) = √3
⇒ tan 20° + tan 40° = √3(1 – tan 20 tan 40)
⇒ tan 20° + tan 40° + √3 tan 20 tan 40) = √3

Question 21.
Prove that \(\frac{\cos 9^{\circ}+\sin 9^{\circ}}{\cos 9^{\circ}-\sin 9^{\circ}}\) = cot 36°. [May ’15(AP); Mar. ’15(AP); Mar. ’11; N.P]
Answer:

= tan (90° – 36°) = cot 36° = RHS

Question 22.
Show that cos 42° + cos 78° + cos 162° = 0.
Answer:
L.H.S = cos 42° + cos 78° + cos 162°
= cos(60° – 18°) + cos (60° + 18°) + cos (180° – 18°)
= 2cos 60° cos 18° – cos 18°
= 2\(\left(\frac{1}{2}\right)\) cos 18° – cos 18° = cos 18° – cos 18°
= 0 = R.H.S

Question 23.
Simplify sin 1140°. cos 390° – cos 780° sin 750°.
Answer:
sin 1140°. cos 390° – cos 780° sin 750°
= sin[3(360) + 60°] cos [360 + 30°] – cos[2(360) + 60°]sin[2 × 360 + 30°]
= sin 60° cos 30° – cos 60° sin 30°
= sin(60° – 30°) = sin 30° = \(\frac{1}{2}\)

Question 24.
If sin(θ + α) = cos(θ + α), then express tan θ in terms of tan α.
Answer:
sin(θ + α) = cos(θ + α)
⇒ tan(θ + α) = 1
⇒ \(\frac{\tan \theta+\tan \alpha}{1-\tan \theta \tan \alpha}\) = 1
⇒ tan θ + tan α = 1 – tan θ tan α
⇒ tan θ + tan α + tan θ tan α = 1
⇒ tan θ[1 + tan α] = 1 – tan α
∴ tan θ = \(\frac{1-\tan \alpha}{1+\tan \alpha}\)

Question 25.
If cos θ = \(\frac{-5}{13}\) and \(\frac{\pi}{2}\) < θ < π find the value of sin 2θ.
Answer:
\(\frac{\pi}{2}\) < θ < π ⇒ sin θ > 0 and cos θ = \(\frac{-5}{13}\)
⇒ sin θ = \(\frac{12}{13}\)
∴ sin 2θ = 2sin θ cos θ
= 2\(\left(\frac{12}{13}\right)\left(-\frac{5}{13}\right)=-\frac{120}{169}\)

Question 26.
Express \(\frac{1-\cos \theta+\sin \theta}{1+\cos \theta+\sin \theta}\) in terms of tan\(\frac{\theta}{2}\).
Answer:

Question 27.
If 0 < θ < \(\frac{\pi}{2}\), show that \(\sqrt{2+\sqrt{2+\sqrt{2+2 \cos 4 \theta}}}\) = 2cos(θ/2). [Mar. ’02]
Answer:

Question 28.
Prove that \(\frac{1}{\sin 10^{\circ}}-\frac{\sqrt{3}}{\cos 10^{\circ}}\) = 4. [Mar. ’18(TS), Mar. ’16(AP), ’03; May ’04]
Answer:

Question 29.
If \(\frac{\sin \alpha}{a}=\frac{\cos \alpha}{b}\), then prove that a sin 2α + b cos 2α = b. [Mar. ’10, ’01; May 05]
Answer:
Given that \(\frac{\sin \alpha}{a}=\frac{\cos \alpha}{b}\)
⇒ b sin α = a cos α
L.H.S = a sin 2α + b cos 2α
= a(2 sinα cos α) + b(1 – 2sin²α)
= 2sin α(a cos α) + b – 2b sin²α
= 2 sin α(b sin α) + b – 2b sin²α
= 2b sin²α + b – 2b sin²α = b

Question 30.
Prove that sin 78° + cos 132° = \(\frac{\sqrt{5}-1}{4}\)
Answer:
sin 78° + cos 132° = sin 78° + cos (90 + 42)
= sin 78° – sin 42°
= 2 cos\(\left(\frac{78^{\circ}+42^{\circ}}{2}\right)\) sin\(\left(\frac{78^{\circ}-42^{\circ}}{2}\right)\)
= 2 cos 60° sin 18° = 2\(\left(-\frac{1}{2}\right)\left(\frac{\sqrt{5}-1}{4}\right)\)
= \(\frac{\sqrt{5}-1}{4}\) = R.H.S

Question 31.
Find the value of sin 34° + cos 64° – cos 4°. [May ’14]
Answer:
sin 34° + cos 64° – cos 4°
= sin 34° + 2sin \(\left(\frac{64+4}{2}\right)\) sin\(\left(\frac{4^{\circ}-64^{\circ}}{2}\right)\)
= sin 34° + 2sin 34° sin(-30°)
= sin 34° + 2sin 34°(-1/2) = 0

Question 32.
Prove that 4(cos 66° + sin 84°) = √3 + \(\sqrt{15}\). [May ’01]
Answer:
4(cos 66° + sin 84°) = 4[cos 66° + sin(90 – 6°)]
= 4[cos 66° + cod 6°]

Question 33.
Prove that cos 48° cos 12° = \(\frac{3+\sqrt{5}}{8}\).
Answer:

Question 34.
Find the period of f(x) = cos\(\left(\frac{4 x+9}{5}\right)\)
Answer:
The function f(x)
= cos x ∀ x ∈ R has the period 2π.
∴ f(x) = cos\(\left(\frac{4 x+9}{5}\right)\) is periodic and period of f is \(\frac{2 \pi}{\frac{4}{5}}=\frac{5 \pi}{2}\)

Question 35.
Find the period of f(x) = tan 5x.
Answer:
The function tan x is periodic with period π.
∴ f(x) = tan 5x is periodic and its period is \(\frac{\pi}{|5|}=\frac{\pi}{5}\)

Question 36.
Find the period of f(x) = |sin x|.
Answer:
The function sin x has period 2π ∀ x ∈ R.
But f(x) = |sin x| is periodic and its period is π.
[∵ f(x + π) = |sin(x + π)| = |- sin x| = sin x]

Question 37.
Find the period of f(x) = tan(x + 4x + 9x + …………. + n²x) (n any positive integer). [B.P. Mar ’15(AP & TS)]
Answer:
Given f(x) = tan(x + 4x + 9x + …………. + n²x)
tan(1 + 2² + 3² + …………… + n²)x

Question 38.
Find a sine function whose period is \(\frac{2}{3}\).
Answer:
\(\frac{2 \pi}{\mathrm{k}}=\frac{2}{3}\) ⇒ 3π = |k|
∴ sin kx = sin(3π x)

Question 39.
Find a cosine function whose period is 7.
Answer:
Let f(x) = cos kx
Period of cos kx = \(\frac{2 \pi}{|k|}\)
∴ \(\frac{2 \pi}{|k|}\) = 7 ⇒ |k| = \(\frac{2 \pi}{|k|}\)
∴ f(x) = cos[\(\frac{2 \pi}{|k|}\). x]

Question 40.
Find the period of cos⁴x.
Answer:

Question 41.
Find the period of 2 sin \(\left(\frac{\pi \mathbf{x}}{4}\right)\) + 3 cos \(\left(\frac{\pi \mathbf{x}}{3}\right)\).
Answer:

Question 42.
Find the minimum and maximum values of f(x) = 3 cos x + 4 sin x.
Answer:
Recall for a cos x + b sin x + c
Max value = c + \(\sqrt{a^2+b^2}\) and Min value
= c – \(\sqrt{a^2+b^2}\)
a = 3, b = 4, c = 0
∴ Max. value = \(\sqrt{9+16}\) =5
Min. value = –\(\sqrt{9+16}\) = – 5

Find the maximum and minimum values of f(x) = 3 sin x – 4 cos x.
Answer:
5, -5.

Question 43.
Find the maximum and minimum values of cos (x + \(\frac{\pi}{3}\)) + 22sin(x + \(\frac{\pi}{3}\)) – 3
Answer:
Let f(x) = cos(x + \(\frac{\pi}{3}\)) + 22sin(x + \(\frac{\pi}{3}\)) – 3
Comparing the given expression with
a sin x + b cos x + c, we get a = 2√2 , b = 1, c = – 3
∴ Maximum value of f(x) is
c + \(\sqrt{(2 \sqrt{2})^2+(1)^2}\) = -3 + \(\sqrt{(2 \sqrt{2})^2+(1)^2}\)
= -3 + \(\sqrt{8+1}\) = -3 + 3 = 0

∴ Minimum value of f(x) is
c – \(\sqrt{a^2+b^2}\) = -3 – \(\sqrt{(2 \sqrt{2})^2+(1)^2}\)
= – 3 – \(\sqrt{8+1}\) = -3 – 3 = -6

Question 44.
Find the range of 13 cos x + 3√3 sin x – 4.
Answer:
Let f(x) = 13 cos x + 3√3 sin x – 4.
a = 3√3, b = 13, c = -4

Find the range of 7 cos x – 24 sin x + 5
Answer:
[-20, 30]

Question 45.
Find the extreme values of cos 2x + cos²x.
Answer:
cos 2x + cos²x = 2cos²x – 1 + cos²x = 3cos²x – 1
and 0 ≤ cos²x ≤ 1
⇒ 0 ≤ 3 cos²x ≤ 3
⇒ -1 ≤ 3 cos²x – 1 ≤ 2

Maximum value = 2 and minimum value = -1
(or) cos 2x + cos²x = cos 2x + \(\left(\frac{1+\cos 2 x}{2}\right)\)
We have -1 ≤ cos 2x ≤ 1 ⇒ -3 ≤ 3cos 2x ≤ 3
-2 ≤ 3 cos 2x + 1 ≤ 4
-1 ≤ \(\frac{3 \cos 2 x+1}{2}\) ≤ 2

Maximum value = 2
Maximum value = -1 (or) a = \(\frac{3}{2}\), b = 0, c = \(\frac{1}{2}\)
Minimum value c – \(\sqrt{a^2+b^2}=\frac{1}{2}-\sqrt{9 / 4}\)
= \(\frac{1}{2}-\frac{3}{2}\) = -1
Maximum value : c + \(\sqrt{a^2+b^2}\) = \(\frac{1}{2}+\frac{3}{2}\) = 2

Question 46.
Find the extreme values of 3 sin²x + 5 cos² x.
Answer:

Question 47.
Sketch the graph of tan x between 0 and \(\frac{\pi}{4}\).
Answer:

Question 48.
Sketch the graph of cos 2x in the interval [0, π]
Answer:

Question 49.
Sketch the graph of sin 2x in the interval (0, π).
Answer:

Question 50.
Sketch the graph of sin x in the interval [-π, + π] taking four values on X – axis.
Answer:

Question 51.
Sketch the graph of cos²x in [0, π].
Answer:

Question 52.
Sketch the region enclosed by y = sin x, y = cos x and X – axis In the Interval [0, π].
Answer: