Students get through Maths 1A Important Questions Inter 1st Year Maths 1A Inverse Trigonometric Functions Important Questions which are most likely to be asked in the exam.

## Intermediate 1st Year Maths 1A Inverse Trigonometric Functions Important Questions

Question 1.
Find the values of the following.
i) sin-1(-$$\frac{1}{2}$$)
Solution:
sin-1(-$$\frac{1}{2}$$) = -sin-1($$\frac{1}{2}$$) = –$$\frac{\pi}{6}$$

ii) cos-1(-$$\frac{\sqrt{3}}{2}$$)
Solution:
cos-1(-$$\frac{\sqrt{3}}{2}$$) = π – cos-1($$\frac{\sqrt{3}}{2}$$)
= π – $$\frac{\pi}{6}$$ = $$\frac{5\pi}{6}$$

iii) tan-1($$\frac{1}{\sqrt{3}}$$)
Solution:
tan-1($$\frac{1}{\sqrt{3}}$$) = $$\frac{\pi}{6}$$

iv) cot-1(-1)
Solution:
cot-1(-1) = π – cot-1 (1) = π – $$\frac{\pi}{4}$$
= $$\frac{3\pi}{4}$$

v) sec-1(-$$\sqrt{2}$$)
Solution:
sec-1(-$$\sqrt{2}$$) = π – sec-1($$\sqrt{2}$$)
= π – $$\frac{\pi}{4}$$ = $$\frac{3\pi}{4}$$

vi) cosec-1($$\frac{2}{\sqrt{3}}$$)
Solution:
cosec-1($$\frac{2}{\sqrt{3}}$$) = sin-1($$\frac{\sqrt{3}}{2}$$) = $$\frac{\pi}{3}$$

Question 2.
Find the values of the following.
i) sin-1(sin $$\frac{4\pi}{3}$$)
Solution:

ii) cos-1(cos $$\frac{4\pi}{3}$$)
Solution:
cos-1(cos $$\frac{4\pi}{3}$$)
= cos-1 (cos (π + $$\frac{\pi}{3}$$))
= cos-1 (-cos $$\frac{\pi}{3}$$)
= π – cos-1 (cos $$\frac{\pi}{3}$$) = π – $$\frac{\pi}{3}$$ = $$\frac{2\pi}{3}$$;
∵ $$\frac{2\pi}{3}$$ ∈ (0, π)

iii) tan-1(tan $$\frac{4\pi}{3}$$)
Solution:
tan-1(tan $$\frac{4\pi}{3}$$) = tan-1(tan(π + $$\frac{\pi}{3}$$))
= tan-1(tan $$\frac{\pi}{3}$$) = $$\frac{\pi}{3}$$;
∵ $$\frac{\pi}{3}$$ ∈ (-$$\frac{\pi}{2}$$, $$\frac{\pi}{2}$$)

Question 3.
Find the values of the following
i) sin(cos-1 $$\frac{5}{13}$$)
Solution:
sin(cos-1 $$\frac{5}{13}$$) = sin (sin-1 $$\frac{12}{13}$$) = $$\frac{12}{13}$$

ii) tan (sec-1 $$\frac{25}{7}$$)
Solution:
tan (sec-1 $$\frac{25}{7}$$) = tan (tan-1 $$\frac{24}{7}$$) = $$\frac{24}{7}$$

iii) cos (tan-1 $$\frac{24}{7}$$)
Solution:
cos (tan-1 $$\frac{24}{7}$$) = cos (cos-1 $$\frac{7}{25}$$) = $$\frac{7}{25}$$

Question 4.
Find the values of the following
i) sin2 (tan-1 $$\frac{3}{4}$$)
Solution:
sin (tan-1 $$\frac{3}{4}$$) = sin (sin-1 $$\frac{3}{5}$$) = $$\frac{3}{5}$$
∴ sin2 (tan-1 $$\frac{3}{4}$$) = ($$\frac{3}{5}$$)2 = $$\frac{9}{25}$$

ii) sin ($$\frac{\pi}{2}$$ – sin-1(-$$\frac{4}{5}$$))
Solution:
sin ($$\frac{\pi}{2}$$ – sin-1(-$$\frac{4}{5}$$)
= sin ($$\frac{\pi}{2}$$ – sin-1($$\frac{4}{5}$$)
= cos (sin-1 $$\frac{4}{5}$$)
= cos (cos-1 $$\frac{3}{5}$$) = $$\frac{3}{5}$$

iii) cos (cos-1(-$$\frac{2}{3}$$) – sin-1($$\frac{2}{3}$$))
Solution:
cos (cos-1(-$$\frac{2}{3}$$) – sin-1($$\frac{2}{3}$$))
= cos (π – cos-1$$\frac{2}{3}$$ – sin-1($$\frac{2}{3}$$))
= cos (π – (cos-1$$\frac{2}{3}$$ + sin-1$$\frac{2}{3}$$))
= cos (π – $$\frac{\pi}{2}$$) = cos ($$\frac{\pi}{2}$$) = 0

iv) sec2(cot-1 3) + cosec2 (tan-1 2)
Solution:
Let cot-1 (3) = α and tan-1 (2) = β
Then cot α = 3 and tan β = 2
⇒ tan α = $$\frac{1}{3}$$ and cot β = $$\frac{1}{2}$$
Now sec2(cot-1 3) + cosec2 (tan-1 2)
= sec2 α + cosec2 β
= (1 + tan2α) + (1 + cot2 β)
= 1 + ($$\frac{1}{3}$$)2 + 1 + ($$\frac{1}{2}$$)2
= 2 + $$\frac{1}{9}$$ + $$\frac{1}{4}$$
= $$\frac{72+4+9}{36}$$ = $$\frac{85}{36}$$

Question 5.
Find the value of cot-1 $$\frac{1}{2}$$ + cot-1 $$\frac{1}{3}$$
Solution:
cot-1 $$\frac{1}{2}$$ + cot-1 $$\frac{1}{3}$$
= tan-1 (2) + tan-1 (3)
∵ x = 3, y = 2, xy > 1
= π + tan-1 ($$\frac{2+3}{1-(2)(3)}$$)
= π + tan-1 ($$\frac{5}{-5}$$)
= π + tan-1 (-1)
= π – $$\frac{\pi}{4}$$ = $$\frac{3\pi}{4}$$

Question 6.
Prove that sin-1 $$\frac{4}{5}$$ + sin-1 $$\frac{7}{25}$$ = sin-1 $$\frac{117}{125}$$ [Mar 16]
Solution:
Method (i):
Let sin-1($$\frac{4}{5}$$) = α and sin-1 $$\frac{7}{25}$$ = β
Then sin α = $$\frac{4}{5}$$ and sin β = $$\frac{7}{25}$$ and α, β ∈ (0, $$\frac{\pi}{2}$$)
So that cos α = $$\frac{3}{5}$$ and cos β = $$\frac{24}{25}$$ and α + β ∈ (0, π)
Now
cos(α + β) = cos α cos β – sin α sin β
= $$\frac{3}{5}$$ . $$\frac{24}{25}$$ – $$\frac{4}{5}$$ . $$\frac{7}{25}$$
= $$\frac{72-28}{125}$$ = $$\frac{44}{125}$$ > 0
⇒ (α + β) ∈ (0, $$\frac{\pi}{2}$$)
Now sin(α + β) = sin α cos β – cos α sin β
= $$\frac{4}{5}$$ . $$\frac{24}{25}$$ – $$\frac{3}{5}$$ . $$\frac{7}{25}$$
= $$\frac{96+21}{125}$$ = $$\frac{117}{125}$$
⇒ (α + β) = sin ($$\frac{117}{125}$$)
∴ sin-1 ($$\frac{4}{5}$$) + sin-1 ($$\frac{7}{25}$$) = sin-1 ($$\frac{117}{125}$$

Method (ii) :
We know that

Question 7.
If x ∈ (-1, 1), prove that 2 tan-1x = tan-1($$\frac{2 x}{1-x^{2}}$$)
Solution:
∵ x ∈ (-1, 1) and let tan-1 x = α
Then tan α = x and $$\frac{-\pi}{4}$$ < $$\frac{\pi}{4}$$
Now tan-1 ($$\frac{2 x}{1-x^{2}}$$) = tan-1 ($$\frac{2 \tan \alpha}{1-\tan ^{2} \alpha}$$)
= tan-1 (tan 2α)
= 2α,
since 2α ∈ (-$$\frac{-\pi}{2}$$, $$\frac{-\pi}{2}$$)
∴ tan-1($$\frac{2 x}{1-x^{2}}$$) = 2 tan-1 (x)

Question 8.
Prove that sin-1 $$\frac{4}{5}$$ + sin-1 $$\frac{5}{13}$$ + sin-1 $$\frac{16}{25}$$ = $$\frac{\pi}{2}$$
Solution:
Let sin-1 $$\frac{4}{5}$$ = α and sin-1 $$\frac{5}{13}$$ = β
Then α, β are acute angles and sin α = $$\frac{4}{5}$$, sin β = $$\frac{5}{13}$$
So that cos α = $$\frac{3}{5}$$ and cos β = $$\frac{12}{13}$$
Now
cos (α + β) = cos α cos β – sin α sin β
= $$\frac{3}{5}$$ . $$\frac{12}{13}$$ – $$\frac{4}{5}$$ . $$\frac{5}{13}$$
= $$\frac{16}{65}$$
∴ α + β = cos-1 ($$\frac{16}{65}$$)
⇒ sin-1 $$\frac{4}{5}$$ + sin-1 $$\frac{5}{13}$$ = cos-1($$\frac{16}{65}$$) __________ (1)
LHS
= (sin-1$$\frac{4}{5}$$ + sin-1 $$\frac{5}{13}$$) + sin-1($$\frac{16}{65}$$)
= cos-1 $$\frac{16}{65}$$ + sin-1 $$\frac{16}{65}$$ = $$\frac{\pi}{2}$$ [By (1)]
LHS = RHS

Question 9.
Prove that cot-1 9 + cosec-1 $$\frac{\sqrt{41}}{4}$$ = $$\frac{\pi}{4}$$
Solution:
Let cot-1 (9) = α and cosec-1 $$\frac{\sqrt{41}}{4}$$ = β
⇒ cot α = 9 and cosec β = $$\frac{\sqrt{41}}{4}$$
⇒ tan α = $$\frac{1}{9}$$ and cot β =
= $$\sqrt{\frac{41}{16}-1}$$ = $$\sqrt{\frac{25}{16}}$$ = $$\frac{5}{4}$$
∴tan α = $$\frac{1}{9}$$ and tan β = $$\frac{4}{5}$$
Now tan(α + β) = $$\frac{\tan \alpha+\tan \beta}{1-\tan \alpha \tan \beta}$$
= $$\frac{\frac{1}{9}+\frac{4}{5}}{1-\left(\frac{1}{9}\right)\left(\frac{4}{5}\right)}$$
= $$\left(\frac{5+36}{45-4}\right)$$ = 1
⇒ tan (α + β) = tan $$\frac{\pi}{4}$$
⇒ α + β = $$\frac{\pi}{4}$$
⇒ cot-1 (9) + cosec-1 ($$\frac{\sqrt{41}}{4}$$) = $$\frac{\pi}{4}$$

Question 10.
Show that cot (sin-1 $$\sqrt{\frac{13}{17}}$$) = sin (tan-1 $$\frac{2}{3}$$)
Solution:

Question 11.
Find the value of tan (2 tan-1($$\frac{1}{5}$$) – $$\frac{\pi}{4}$$)
Solution:

Question 12.
Prove that sin-1 $$\frac{4}{5}$$ + 2 tan-1 $$\frac{1}{3}$$ = $$\frac{\pi}{2}$$
Solution:

Question 13.
Prove that cos (2 tan-1 $$\frac{1}{7}$$) = sin (4 tan-1 $$\frac{1}{3}$$)
Solution:
Let tan-1 $$\frac{1}{7}$$) = α and tan-1 $$\frac{1}{3}$$ = β
Then tan α = $$\frac{1}{7}$$ and tan β = $$\frac{1}{3}$$
Now

Question 14.
If sin-1 x + sin-1y + sin-1z = π, then prove that x4 + y4 + z4 + 4x2y2z2 = 2 (x2y2 + y2z2 + z2x2)
Solution:
Let sin-1x = A, sin-1 y = B and sin-1(z) = c
then A + B + C = π ________(1) and
sin A = x, sin B = y and sin C = z
Now A + B = π – C’
= cos (A + B) = cos( π – C’)
= cos A cos B – sin A sin B = -cos C
⇒ $$\sqrt{1-x^{2}} \sqrt{1-y^{2}}$$ – xy = – $$\sqrt{1-z^{2}}$$
⇒ $$\sqrt{1-x^{2}} \sqrt{1-y^{2}}$$ = xy – $$\sqrt{1-z^{2}}$$
On squaring both sides we get
(1 – x2)(1 – y2) = x2y2 + (1 – z2) – 2xy $$\sqrt{1-z^{2}}$$
⇒ 2xy$$\sqrt{1-z^{2}}$$ = x2 + y2 – z2
Again on squaring both sides, we get
(2xy $$\sqrt{1-z^{2}}$$)2 (x2 + y2 – z2)2
⇒ 4x2y2(1 – z2) = x4 + y4 + z4 + 2x2y2 – 2y2z2 – 2z2x2
⇒ 4x2y2 – 4x2y2z2 = x4 + y4 + z4 + 2x2y2 – 2y2z2 — 2z2x2
⇒ x4 + y4 + z4 + 4x2y2z2 = 2x2y2 + 2y2z2 + 2z2x2

Question 15.
If cos-1$$\frac{p}{a}$$ + cos-1$$\frac{q}{b}$$ = α, then prove that $$\frac{p^{2}}{a^{2}}$$ – $$\frac{2 p q}{a b}$$ cos α + $$\frac{q^{2}}{b^{2}}$$ = sin2 α
Solution:
Let cos-1$$\frac{p}{a}$$ = α and cos-1$$\frac{q}{b}$$ = β
then cosα = $$\frac{p}{a}$$ and cos β = $$\frac{q}{b}$$ and
A + B = α (given)
Now
cos α = cos(A + B)
= cosA cosB – sinA sinB

Question 16.
Solve are sin($$\frac{5}{x}$$) + arc sin $$\frac{12}{x}$$ = $$\frac{\pi}{2}$$; (x > 0)
Solution:
Given that
sin-1 ($$\frac{5}{x}$$) + sin-1 $$\frac{12}{x}$$ = $$\frac{\pi}{2}$$; x > 0
Let sin-1 $$\frac{5}{x}$$ = α and sin-1 $$\frac{12}{x}$$ = β
then sin α = $$\frac{5}{x}$$ and sin β = $$\frac{12}{x}$$, x > 0
Now α + β = $$\frac{\pi}{2}$$
⇒ α = $$\frac{\pi}{2}$$ – β
sin α = sin($$\frac{\pi}{2}$$ – β) ⇒ sin α = cos β
= $$\frac{5}{x}$$ = $$\sqrt{1-\left(\frac{12}{x}\right)^{2}}$$
⇒ $$\frac{25}{x^{2}}$$ = 1 – $$\frac{144}{x^{2}}$$
⇒ $$\frac{169}{x^{2}}$$ = 1 ⇒ x2 = 169 ⇒ x = ± 13
⇒ x = 13 (∵ x > 0)

Question 17.
Solve sin-1 ($$\frac{3x}{5}$$) + sin-1 ($$\frac{4x}{5}$$) = sin-1(x)
Solution:
Let sin-1 ($$\frac{3x}{5}$$) = α, sin-1 ($$\frac{4x}{5}$$) = β and sin-1(x) = γ
Then sin α = $$\frac{3x}{5}$$, sin β = $$\frac{3x}{5}$$ and sin γ = x
⇒ cos α = $$\sqrt{1-\frac{9 x^{2}}{25}}$$, cos β = $$\sqrt{1-\frac{16 x^{2}}{25}}$$ and cos γ = $$\sqrt{1-x^{2}}$$
Now α + β = γ
⇒ sin (α + β) = sin γ
⇒ sinα cos β + cos α sin β = sin γ

Squaring on both sides
16(25 – 9x2) = 625 – 150$$\sqrt{25-16 x^{2}}$$ + 9(25 – 16x2)
⇒ 400 – 144x2 = 625 – 150$$\sqrt{25-16 x^{2}}$$ + 225 – 144x2
⇒ 150$$\sqrt{25-16 x^{2}}$$ = 225 + 225
⇒ $$\sqrt{25-16 x^{2}}$$ = 3
⇒ 25 – 16x2 = 9
⇒ 16x2 = 16 ⇒ x = ± 1
∴ x = 0, + 1, – 1.
All these values of x satisfy the given equation.

Question 18.
Solve sin<sup>-1</sup> x + sin<sup>-1</sup> 2x = $$\frac{\pi}{3}$$
Solution:
Given that sin<sup>-1</sup> x + sin<sup>-1</sup> 2x = $$\frac{\pi}{3}$$
⇒ cos (sin<sup>-1</sup>x + sin<sup>-1</sup> 2x) = cos($$\frac{\pi}{3}$$)

But when x = –$$\frac{\sqrt{3}}{2 \sqrt{7}}$$,. both sin<sup>-1</sup>x and sin<sup>-1</sup>(2x) are negative. The given equation does not satisfy.
Hence x = $$\frac{\sqrt{3}}{2 \sqrt{7}}$$ is the only solution.

Question 19.
If sin [2 cos-1 {cot (2 tan-1 x)}] = 0, find x
Solution:
sin [2cos-1 {cot (2 tan-1 x)}] = 0
⇔ 2 cos-1 [cot (2tan-1 x)] = 0 or π or 2π
(since the range of cos-1 is (0, π)
⇔ cos-1 [cot (2 tan-1 x)] = 0 or $$\frac{\pi}{2}$$ or π
⇒ cot (2 tan-1 x) = 1 or 0 or -1
⇒ 2 tan-1 x = ±$$\frac{\pi}{4}$$ or ±$$\frac{\pi}{2}$$ (or) ±$$\frac{3\pi}{4}$$
∴ tan-1 (x) = ±$$\frac{\pi}{8}$$ (or) ±$$\frac{\pi}{4}$$ (or) ± $$\frac{3\pi}{8}$$
x = ± ($$\sqrt{2}$$ – 1) (or) ±1 (or) ± ($$\sqrt{2}$$ + 1)

Question 20.
Prove that cos-1 [tan-1 (sin (cot-1x)}] = $$\sqrt{\frac{x^{2}+1}{x^{2}+2}}$$
Solution:
Let cot-1 (x) = θ,
then cot θ = x and 0 < x < π