Chapter 2 Inverse Trigonometric Functions Exercise 2 Question Answer Elements of Mathematics Class 12 CHSE Odisha

Chapter 2 Inverse Trigonometric Functions Exercise 2 CHSE Odisha Class 12 Math Solutions | Elements of Mathematics 

Question 1.

Fill in the blanks choosing correct answer from the brackets:

(i) If A = tan^-1 x, then the value of sin 2A = ________. ($\frac{2 x}{1-x^2}$, $\frac{2 x}{\sqrt{1-x^2}}$, $\frac{2 x}{1+x^2}$)

Solution:

$\frac{2 x}{1+x^2}$

(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 –

(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

Solution:

True

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

Solution:

True

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

Solution:

False

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

Solution:

True

(vi) tan^-1 (tan 3) = 3

Solution:

False

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

Solution:

False

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

Solution:

True

(ix) 3 tan^-1 3 = tan^-1

Solution:

False

(x) tan^-1 2 + tan^-1 3 = –

Solution:

False

(xi) 2 sin^-1 $\frac{4}{5}$ = sin^-1

Solution:

False

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

Solution:

True

Question 3.

Express the value of the following 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

Solution:

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

Solution:

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

Solution:

(viii) sin cos^-1 tan sec √2

Solution:

sin cos^-1 tan sec √2

= sin cos^-1 tan sec

= sin cos^-1 1 = sin 0 = 0

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

Solution:

(x) tan

Solution:

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

Solution:

(xii) tan^-1

Solution:

Question 4.

Prove the following statements:

(i) sin^-1 $\frac{3}{5}$ + sin^-1 $\frac{8}{17}$ = cos^-1

Solution:

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

Solution:

(iii) tan^-1 $\frac{1}{7}$ + tan^-1 $\frac{1}{13}$ = tan^-1

Solution:

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

Solution:

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

Solution:

Question 5.

Prove the following statements:

(i) cot^-1 9 + cosec^-1 $\frac{\sqrt{41}}{4}$ =

Solution:

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

Solution:

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

Solution:

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

Solution:

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

Solution:

(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

= 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:

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

Solution:

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

Solution:

tan^-1 ($\frac{a-b}{1+a b}$) + tan^-1 ($\frac{b-c}{1+b c}$)

= tan^-1 a – tan^-1 b + tan^-1 b – tan^-1 c

= tan^-1 a – tan^-1 c.

(iv) cot^-1 $\frac{p q+1}{p-q}$ + cot^-1 $\frac{q r+1}{q-r}$ + cot^-1 $\frac{r p+1}{r-p}$ = 0

Solution:

(v)

Solution:

Question 7.

Prove the following statements:

(i) tan^-1 $\frac{2 a-b}{b \sqrt{3}}$ + tan^-1 $\frac{2 b-a}{a \sqrt{3}}$ =

Solution:

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

Solution:

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

Solution:

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

Solution:

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

Solution:

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

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

Solution:

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

Solution:

or, 1 – xy – yz – zx = 0

⇒ xy + yz + zx = 1

(iv) If r² = x² +y² + z², Prove that tan^-1 $\frac{y z}{x r}$ + tan^-1 $\frac{z x}{y r}$ + tan^-1 $\frac{x y}{z r}$ =

Solution:

(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

Question 9.

Solve

(i) cos (2 sin^-1 x) = 1/9

Solution:

(ii) sin^-1 x + sin^-1 (1 – x) =

Solution:

sin^-1 x + sin^-1 (1 – x) =

or, sin^-1 (1 – x) = $\frac{\pi}{2}$ – sin^-1 x = cos^-1 x

or, sin^-1 (1 – x) = sin^-1

or, 1 – x =

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 =

Solution:

sin^-1 (1 – x) – 2 sin^-1 x =

⇒ – 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 $\frac{x}{2}$ =

Solution:

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

Solution:

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

Solution:

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

Solution:

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

Solution:

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

Solution:

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

Solution:

(xi) sin^-1 y – cos^-1 x = cos^-1

Solution:

(xii) sin^-1 2x + sin^-1 x =

Solution:

Question 10.

Rectify the error if any in the following:

sin^-1 $\frac{4}{5}$ + sin^-1 $\frac{12}{13}$ + sin^-1

Solution:

Question 11.

Prove that:

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

Solution:

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

Solution:

(iii) tan^-1 $\sqrt{\frac{x r}{y z}}$ + tan^-1 $\sqrt{\frac{y r}{y x}}$ + tan^-1 $\sqrt{\frac{z r}{x y}}$ = π where r = x + y +z.

Solution:

Question 12.

(i) If cos^-1 ($\frac{x}{a}$) + cos^-1 ($\frac{y}{b}$) = Θ, prove that $\frac{x^2}{a^2}$ – $\frac{2 x}{a b}$ 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 ($\frac{c^2}{a b}$) 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 $\frac{x^2}{a^2}$ + $\frac{2 x y}{a b}$ 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 $\frac{2 x-1}{2 x+1}$ = tan^-1

Solution:

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

Solution:

(iii) cos^-1 $\left(x+\frac{1}{2}\right)$ + cos^-1 x+ cos^-1 $\left(x-\frac{1}{2}\right)$ =

Solution:

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

Solution: