CHSE Odisha Class 12 Math Solutions Chapter 2 Inverse Trigonometric Functions Exercise 2

CHSE Odisha Class 12 Math Solutions Chapter 2 Inverse Trigonometric Functions Exercise 2

Question 1.
Fill in the blanks choosing correct answer from the brackets:
(i) If A = tan^-1 x, then the value of sin 2A = ________. ($\frac{2x}{1-x^2}$, $\frac{2x}{\sqrt{1-x^2}}$, $\frac{2x}{1+x^2}$)
Solution:
$\frac{2x}{1+x^2}$

(ii) If the value of sin^-1 x = $\frac{\pi }{5}$ for some x ∈ (-1, 1) then the value of cos^-1 x is ________. ($\frac{3\pi }{10}$, $\frac{5\pi }{10}$,$\frac{3\pi }{10}$)
Solution:
$\frac{3\pi }{10}$

(iii) The value of tan^-1 x (2cos$\frac{\pi }{3}$) is ________. (1, $\frac{\pi }{4}$, $\frac{\pi }{3}$)
Solution:
$\frac{\pi }{4}$

(iv) If x + y = 4, xy = 1, then tan^-1 x + tan^-1 y = ________. ($\frac{3\pi }{4}$, $\frac{\pi }{4}$, $\frac{\pi }{3}$)
Solution:
$\frac{\pi }{2}$

(v) The value of cot^-1 2 + tan^-1 $\frac{1}{3}$ = ________. ($\frac{\pi }{4}$, 1, $\frac{\pi }{2}$)
Solution:
$\frac{\pi }{4}$

(vi) The principal value of sin^-1 (sin $\frac{2\pi }{3}$) is ________. ($\frac{2\pi }{3}$, $\frac{\pi }{3}$, $\frac{4\pi }{3}$)
Solution:
$\frac{\pi }{3}$

(vii) If sin^-1 $\frac{x}{5}$ + cosec^-1 $\frac{5}{4}$ = $\frac{\pi }{2}$, then the value of x = ________. (2, 3, 4)
Solution:
x = 3

(viii) The value of sin (tan^-1 x + tan^-1 $\frac{1}{x}$), x > 0 = ________. (0, 1, 1/2)
Solution:
1

(ix) cot^-1 $\left[\frac{\sqrt{1-\sin x}+\sqrt{1+\sin x}}{\sqrt{1-\sin x}-\sqrt{1+\sin x}}\right]$ = ________. (2π – $\frac{x}{2}$, $\frac{x}{2}$, π – $\frac{x}{2}$)
Solution:
π – $\frac{x}{2}$

(x) 2sin^-1 $\frac{4}{5}$ + sin^-1 $\frac{24}{25}$ = ________. (π, -π, 0)
Solution:
π

(xi) if Θ = cos^-1 x + sin^-1 x – tan^-1 x, x ≥ 0, then the smallest interval in which Θ lies is ________. [($\frac{\pi }{2}$, $\frac{3\pi }{2}$), [0, $\frac{\pi }{2}$), (0, $\frac{\pi }{2}$])
Solution:
(0, $\frac{\pi }{2}$]

(xii) sec² (tan^-1 2) + cosec² (cot^-1 3) = ________. (16, 14, 15)
Solution:
15

Question 2.
Write whether the following statements are true or false.
(i) sin^-1 $\frac{1}{x}$ cosec^-1 x = 1
Solution:
False

(ii) cos^-1 $\frac{4}{5}$ + tan^-1 $\frac{2}{3}$ = tan^-1 $\frac{17}{6}$
Solution:
True

(iii) tan^-1 $\frac{4}{3}$ + cot^-1 ($\frac{-3}{4}$) = π
Solution:
True

(iv) sec^-1 $\frac{1}{2}$ + cosec^-1 $\frac{1}{2}$ = $\frac{\pi }{2}$
Solution:
False

(v) sec^-1 (-$\frac{7}{5}$) = π – cos^-1 $\frac{5}{7}$
Solution:
True

(vi) tan^-1 (tan 3) = 3
Solution:
False

(vii) The principal value of tan^-1 (tan $\frac{3\pi }{4}$) is $\frac{3\pi }{4}$
Solution:
False

(viii) cot^-1 (-√3) is in the second quadrant.
Solution:
True

(ix) 3 tan^-1 3 = tan^-1 $\frac{9}{13}$
Solution:
False

(x) tan^-1 2 + tan^-1 3 = – $\frac{\pi }{4}$
Solution:
False

(xi) 2 sin^-1 $\frac{4}{5}$ = sin^-1 $\frac{24}{25}$
Solution:
False

(xii) The equation tan^-1 (cotx) = 2x has exactly two real solutions.
Solution:
True

Question 3.
Express the value of the folowing in simplest form.
(i) sin (2 sin^-1 0.6)
Solution:
sin (2 sin^-1 0.6)

(ii) tan ($\frac{\pi }{4}$ + 2 cot^-1 3)
Solution

:

(iii) cos (2 sin^-1 x)

Solution:

(iv) tan (cos^-1 x)
Solution:

(v) tan^-1 ($\frac{x}{y}$) – tan^-1 $\frac{x-y}{x+y}$
Solution:

(vi) cosec (cos^-1 $\frac{3}{5}$ + cos^-1 $\frac{4}{5}$)
Solution:

(vii) sin^-1 $\frac{1}{\sqrt{5}}$ + cos^-1 $\frac{3}{\sqrt{10}}$
Solution:

(viii) sin cos^-1 tan sec √2
Solution:
sin cos^-1 tan sec √2
= sin cos^-1 tan sec $\frac{\pi }{4}$
= sin cos^-1 1 = sin 0 = 0

(ix) sin (2 tan^-1 $\sqrt{\frac{1-x}{1+x}}$)
Solution:

(x) tan $\{\frac{1}{2}{\sin }^{-1}\frac{2x}{1+x^2}+\frac{1}{2}{\cos }^{-1}\frac{1-y^2}{1+y^2}\}$
Solution:

(xi) sin cot^-1 cos tan^-1 x.
Solution:

(xii) tan^-1 $(x+\sqrt{1+x^2})$
Solution:

Question 4.
Prove the following statements:
(i) sin^-1 $\frac{3}{5}$ + sin^-1 $\frac{8}{17}$ = cos^-1 $\frac{36}{85}$
Solution:

L.H.S = sin^-1 $\frac{3}{5}$ + sin^-1 $\frac{8}{17}$

(ii) sin^-1 $\frac{3}{5}$ + cos^-1 $\frac{12}{13}$ = cos^-1 $\frac{33}{65}$
Solution:

L.H.S = sin^-1 $\frac{3}{5}$ + cos^-1 $\frac{12}{13}$

Or

Let α = sin^-1 (³/₅) and β = cos^-1 (¹²/₁₃).

Find cos α and sin β:
cos α = ⁴/₅
sin β = ⁵/₁₃

Calculate cos (α + β) using:
cos (α + β) = cos α cos β – sin α sin β
cos (α + β) = ⁴/₅ × ¹²/₁₃ – ³/₅ × ⁵/₁₃ = ⁴⁸/₆₅ – ¹⁵/₆₅ = ³³/₆₅

Thus:
sin^-1 (³/₅) + cos^-1 (¹²/₁₃) = cos^-1 (³³/₆₅)

(iii) tan^-1 $\frac{1}{7}$ + tan^-1 $\frac{1}{13}$ = tan^-1 $\frac{2}{9}$
Solution:
L.H.S = tan^-1 $\frac{1}{7}$ + tan^-1 $\frac{1}{13}$

(iv) tan^-1 $\frac{1}{2}$ + tan^-1 $\frac{1}{5}$ + tan^-1 $\frac{1}{8}$ = $\frac{\pi }{4}$
Solution:

L.H.S = tan^-1 $\frac{1}{2}$ + tan^-1 $\frac{1}{5}$ + tan^-1 $\frac{1}{8}$

(v) tan ( 2tan^-1 $\frac{1}{5}$ – $\frac{\pi }{4}$ ) + $\frac{7}{17}$ = 0
Solution:

tan ( 2tan^-1 $\frac{1}{5}$ – $\frac{\pi }{4}$ ) + $\frac{7}{17}$

Question 5.
Prove the following statements:
(i) cot^-1 9 + cosec^-1 $\frac{\sqrt{41}}{4}$ = $\frac{\pi }{4}$
Solution: L.H.S= cot^-1 9 + cosec^-1 $\frac{\sqrt{41}}{4}$ 

(ii) sin^-1 $\frac{4}{5}$ + 2 tan^-1 $\frac{1}{3}$ = $\frac{\pi }{2}$
Solution:

(iii) 4 tan^-1 $\frac{1}{5}$ – tan^-1 $\frac{1}{70}$ + tan^-1 $\frac{1}{99}$ = $\frac{\pi }{4}$
Solution:

L.H.S

= 4 tan^-1 $\frac{1}{5}$ – tan^-1 $\frac{1}{70}$ + tan^-1 $\frac{1}{99}$

(iv) 2 tan^-1 $\frac{1}{5}$ + sec^-1 $\frac{5\sqrt{2}}{7}$ + 2 tan^-1 $\frac{1}{8}$ = $\frac{\pi }{4}$
Solution:

2 tan^-1 $\frac{1}{5}$ + sec^-1 $\frac{5\sqrt{2}}{7}$ + 2 tan^-1 $\frac{1}{8}$

(v) cos^-1 $\frac{12}{13}$ + 2 cos^-1 $\sqrt{\frac{64}{65}}$ + cos^-1 $\sqrt{\frac{49}{50}}$ = cos^-1 $\frac{1}{\sqrt{2}}$
Solution:

cos^-1 $\frac{12}{13}$ + 2 cos^-1 $\sqrt{\frac{64}{65}}$ + cos^-1 $\sqrt{\frac{49}{50}}$

(vi) tan² cos^-1 $\frac{1}{\sqrt{3}}$ + cot² sin^-1 $\frac{1}{\sqrt{5}}$ = 6
Solution:
tan² cos^-1 $\frac{1}{\sqrt{3}}$ + cot² sin^-1 $\frac{1}{\sqrt{5}}$
= tan² tan^-1 √2 + cot² cot^-1 (2)
= 2 + 4 = 6

(vii) cos tan^-1 cot sin^-1 x = x.
Solution.

Question 6.
Prove the following statements:
(i) cot^-1 (tan 2x) + cot^-1 (- tan 2x) = π
Solution:
L.H.S = cot^-1 (tan 2x) + cot^-1 (- tan 2x)

(ii) tan^-1 x + cot^-1 (x + 1) = tan^-1 (x² + x + 1)
Solution:

L.H.S = tan^-1 x + cot^-1 (x + 1)

(iii) tan^-1 ($\frac{a-b}{1+ab}$) + tan^-1 ($\frac{b-c}{1+bc}$) = tan^-1 a – tan^-1 c.
Solution:
tan^-1 ($\frac{a-b}{1+ab}$) + tan^-1 ($\frac{b-c}{1+bc}$)
= tan^-1 a – tan^-1 b + tan^-1 b – tan^-1 c
= tan^-1 a – tan^-1 c.

(iv) cot^-1 $\frac{pq+1}{p-q}$ + cot^-1 $\frac{qr+1}{q-r}$ + cot^-1 $\frac{rp+1}{r-p}$ = 0
Solution:

L.H.S = cot^-1 $\frac{pq+1}{p-q}$ + cot^-1 $\frac{qr+1}{q-r}$ + cot^-1 $\frac{rp+1}{r-p}$

(v)

Solution:

Question 7.
Prove the following statements:
(i) tan^-1 $\frac{2a-b}{b\sqrt{3}}$ + tan^-1 $\frac{2b-a}{a\sqrt{3}}$ = $\frac{\pi }{3}$
Solution:

L.H.S = tan^-1 $\frac{2a-b}{b\sqrt{3}}$ + tan^-1 $\frac{2b-a}{a\sqrt{3}}$

(ii) tan^-1 $\frac{1}{x+y}$ + tan^-1 $\frac{y}{x^2+xy+1}$ = tan^-1 $\frac{1}{x}$
Solution:

(iii) sin^-1 $\sqrt{\frac{x-q}{p-q}}$ = cos^-1 $\sqrt{\frac{p-x}{p-q}}$ = cot^-1 $\sqrt{\frac{p-x}{x-q}}$
Solution:

(iv) sin² (sin^-1 x + sin^-1 y + sin^-1 z) = cos² (cos^-1 x + cos^-1 y + cos^-1 z)
Solution:

L.H.S = sin² (sin^-1 x + sin^-1 y + sin^-1 z)

(v) tan (tan^-1 x + tan^-1 y + tan^-1 z) = cot (cot^-1 x + cot^-1 y + cot^-1 z)
Solution:

L.H.S = tan (tan^-1 x + tan^-1 y + tan^-1 z)

Question 8.
(i) If sin^-1 x + sin^-1 y + sin^-1 z = π, show that x$\sqrt{1-x^2}$ + x$\sqrt{1-y^2}$ + x$\sqrt{1-z^2}$ = 2xyz
Solution:
Let sin^-1 x = α, sin^-1 y = β, sin^-1 z = γ
∴ α + β + γ = π
∴ x = sin α, y = sin β, z = sin γ
or, α + β = π – γ
or, sin(α + β) = sin(π – γ) = sin γ
and cos(α + β) = cos(π – γ) = – cos γ

L.H.S = x$\sqrt{1-x^2}$ + x$\sqrt{1-y^2}$ + x$\sqrt{1-z^2}$

(ii) tan^-1 x + tan^-1 y + tan^-1 z = π show that x + y + z = xyz.
Solution:

Given , tan^-1 x + tan^-1 y + tan^-1 z = π

(iii) tan^-1 x + tan^-1 y + tan^-1 z = $\frac{\pi }{2}$. Show that xy + yz + zx = 1
Solution:

Given ,tan^-1 x + tan^-1 y + tan^-1 z = $\frac{\pi }{2}$

or, 1 – xy – yz – zx = 0
⇒ xy + yz + zx = 1

(iv) If r² = x² +y² + z², Prove that tan^-1 $\frac{yz}{xr}$ + tan^-1 $\frac{zx}{yr}$ + tan^-1 $\frac{xy}{zr}$ = $\frac{\pi }{2}$
Solution:

L.H.S = tan^-1 $\frac{yz}{xr}$ + tan^-1 $\frac{zx}{yr}$ + tan^-1 $\frac{xy}{zr}$

(v) In a triangle ABC if m∠A = 90°, prove that tan^-1 $\frac{b}{a+c}$ + tan^-1 $\frac{c}{a+b}$ = $\frac{\pi }{4}$. where a, b, and c are sides of the triangle.
Solution:
L.H.S. tan^-1 $\frac{b}{a+c}$ + tan^-1 $\frac{c}{a+b}$

Question 9.
Solve
(i) cos (2 sin^-1 x) = 1/9
Solution:

(ii) sin^-1 x + sin^-1 (1 – x) = $\frac{\pi }{2}$
Solution:
sin^-1 x + sin^-1 (1 – x) = $\frac{\pi }{2}$
or, sin^-1 (1 – x) = $\frac{\pi }{2}$ – sin^-1 x = cos^-1 x
or, sin^-1 (1 – x) = sin^-1 $\sqrt{1-x^2}$
or, 1 – x = $\sqrt{1-x^2}$
or, 1 + x² – 2x = 1 – x²
or, 2x² – 2x = 0
or, 2x (x – 1) = 0
∴ x = 0 or, 1

(iii) sin^-1 (1 – x) – 2 sin^-1 x = $\frac{\pi }{2}$
Solution:
sin^-1 (1 – x) – 2 sin^-1 x = $\frac{\pi }{2}$
⇒ – 2 sin^-1 x = $\frac{\pi }{2}$ – sin^-1 (1 – x)
⇒ cos^-1 (1 – x)
⇒ cos (– 2 sin^-1 x) = 1 – x ….. (1)
Let sin^-1 Θ ⇒ sin Θ
Now cos (– 2 sin^-1 x) = cos (-2Θ)
= cos 2Θ = 1 – 2 sin² Θ = 1 – 2x²
Using in (1) we get
1 – 2x² = 1 – x
⇒ 2x² – x = 0 ⇒ x (2x – 1) = 0
⇒ x = 0, ½, But x = ½ does not
Satisfy the given equation, Thus x = 0.

(iv) cos^-1 x + sin^-1 $\frac{x}{2}$ = $\frac{\pi }{6}$
Solution:

(v) tan^-1 $\frac{x-1}{x-2}$ + tan^-1 $\frac{x+1}{x+2}$ = $\frac{\pi }{4}$
Solution:

tan^-1 $\frac{x-1}{x-2}$ + tan^-1 $\frac{x+1}{x+2}$ = $\frac{\pi }{4}$

(vi) tan^-1 $\frac{1}{2x+1}$ + tan^-1 $\frac{1}{4x+1}$ = tan^-1 $\frac{2}{x^2}$
Solution:

tan^-1 $\frac{1}{2x+1}$ + tan^-1 $\frac{1}{4x+1}$ = tan^-1 $\frac{2}{x^2}$

(vii) 3 sin^-1 $\frac{2x}{1+x^2}$ – 4 cos^-1 $\frac{1-x^2}{1+x^2}$ + 2 tan^-1 $\frac{2x}{1-x^2}$ = $\frac{\pi }{3}$
Solution:

3 sin^-1 $\frac{2x}{1+x^2}$ – 4 cos^-1 $\frac{1-x^2}{1+x^2}$ + 2 tan^-1 $\frac{2x}{1-x^2}$ = $\frac{\pi }{3}$

(viii) cot^-1 $\frac{1}{x-1}$ + cot^-1 $\frac{1}{x}$ + cot^-1 $\frac{1}{x+1}$ = cot^-1 $\frac{1}{3x}$
Solution:

cot^-1 $\frac{1}{x-1}$ + cot^-1 $\frac{1}{x}$ + cot^-1 $\frac{1}{x+1}$ = cot^-1 $\frac{1}{3x}$

(ix) cot^-1 $\frac{1-x^2}{2x}$ = cosec^-1 $\frac{1+a^2}{2a}$ – sec^-1 $\frac{1+b^2}{1-b^2}$
Solution:

cot^-1 $\frac{1-x^2}{2x}$ = cosec^-1 $\frac{1+a^2}{2a}$ – sec^-1 $\frac{1+b^2}{1-b^2}$

(x) sin^-1 $\left(\frac{2a}{1+a^2}\right)$ + sin^-1 $\left(\frac{2b}{1+b^2}\right)$ = 2 tan^-1 x
Solution:

sin^-1 $\left(\frac{2a}{1+a^2}\right)$ + sin^-1 $\left(\frac{2b}{1+b^2}\right)$ = 2 tan^-1 x

(xi) sin^-1 y – cos^-1 x = cos^-1 $\frac{\sqrt{3}}{2}$
Solution:

sin^-1 y – cos^-1 x = cos^-1 $\frac{\sqrt{3}}{2}$

(xii) sin^-1 2x + sin^-1 x = $\frac{\pi }{3}$
Solution:

sin^-1 2x + sin^-1 x = $\frac{\pi }{3}$

Question 10.
Rectify the error ifany in the following:
sin^-1 $\frac{4}{5}$ + sin^-1 $\frac{12}{13}$ + sin^-1 $\frac{33}{65}$
Solution:

Question 11.
Prove that:
(i) cos^-1 $\left(\frac{b+a\cos x}{a+b\cos x}\right)$ = 2 tan^-1 $(\sqrt{\frac{a-b}{a+b}}\tan \frac{x}{2})$
Solution:

(ii) tan $(\frac{\pi }{4}+\frac{1}{2}{\cos }^{-1}\frac{a}{b})$ + tan $(\frac{\pi }{4}-\frac{1}{2}{\cos }^{-1}\frac{a}{b})$
Solution

(iii) tan^-1 $\sqrt{\frac{xr}{yz}}$ + tan^-1 $\sqrt{\frac{yr}{yx}}$ + tan^-1 $\sqrt{\frac{zr}{xy}}$ = π where r = x + y +z.
Solution:

Question 12.
(i) If cos^-1 ($\frac{x}{a}$) + cos^-1 ($\frac{y}{b}$) = Θ, prove that $\frac{x^2}{a^2}$ – $\frac{2x}{ab}$ cos Θ + $\frac{y^2}{b^2}$ = sin² Θ.
Solution:

(ii) If cos^-1 ($\frac{x}{y}$) + cos^-1 ($\frac{y}{3}$) = Θ, prove that 9x² – 12xy cos Θ + 4y² = 36 sin² Θ.
Solution:

(iii) If sin^-1 ($\frac{x}{a}$) + sin^-1 ($\frac{y}{b}$) = sin^-1 ($\frac{c^2}{ab}$) prove that b²x² + 2xy $\sqrt{a^2{b}^2-c^4}$ a²y² = c²
Solution:

(iv) If sin^-1 ($\frac{x}{a}$) + sin^-1 ($\frac{y}{b}$) = α prove that $\frac{x^2}{a^2}$ + $\frac{2xy}{ab}$ cos α + $\frac{y^2}{b^2}$ = sin² α
Solution:

(v) If sin^-1 x + sin^-1 y + sin^-1 z = π prove that x² + y² + z² + 4x²y²z² = 2 ( x²y² + y²z² + z²x² )
Solution:

Question 13.
Solve the following equations:
(i) tan^-1 $\frac{x-1}{x+1}$ + tan^-1 $\frac{2x-1}{2x+1}$ = tan^-1 $\frac{23}{36}$
Solution:

(ii) tan^-1 $\frac{1}{3}$ + tan^-1 $\frac{1}{5}$ + tan^-1 $\frac{1}{7}$ + tan^-1 x = $\frac{\pi }{4}$
Solution:

(iii) cos^-1 $(x+\frac{1}{2})$ + cos^-1 x+ cos^-1 $(x-\frac{1}{2})$ = $\frac{3\pi }{2}$
Solution:

(iv) 3tan^-1 $\frac{1}{2+\sqrt{3}}$ – tan^-1 $\frac{1}{x}$ = tan^-1 $\frac{1}{3}$
Solution: