# MP Board Class 12th Maths Important Questions Chapter 2 Inverse Trigonometric Functions

## ΨMP Board Class 12th Maths Important Questions Chapter 2 Inverse Trigonometric Functions

### Inverse Trigonometric Functions Important Questions

Inverse Trigonometric Functions Objective Type Questions:

Question 1.

Question 1.
If sin-1x – cos -1 x = $$\frac { \pi }{ 6 }$$, then the value of x is equal to:
(a) $$\frac{1}{2}$$
(b) $$\frac { \sqrt { 3 } }{ 2 }$$
(c) $$\frac{-1}{2}$$
(d) None of these
(a) $$\frac{1}{2}$$

Question 2.
If tan-13 + tan-1 8, then the value of x is equal to:
(a) $$\frac { \pi }{ 2 }$$
(b) $$\frac { \pi }{ 3 }$$
(c) $$\frac { \pi }{ 4 }$$
(d) $$\frac { -3\pi }{ 4 }$$
(b) $$\frac { \pi }{ 3 }$$

Question 3.
tan-1 $$\frac{x}{y}$$ + tan-1 $$\frac{x-y}{x+y}$$ is equal to:
(a) $$\frac { \pi }{ 2 }$$
(b) $$\frac { \pi }{ 3 }$$
(c) $$\frac { \pi }{ 4 }$$
(d) $$\frac { -3\pi }{ 4 }$$
(c) $$\frac { \pi }{ 4 }$$

Question 4.
The value of 2 tan-1 {cosec (tan-1 x ) – tan (cot-1 x)} is equal to:
(a) cot-1x
(b) cot-1$$\frac{1}{x}$$
(c) tan-1x
(d) tan-1$$\frac{1}{x}$$
(c) tan-1x

Question 5.
The value of tan{cos-1$$\frac { 1 }{ 5\sqrt { 2 } }$$ – sin-1 $$\frac { 4 }{ \sqrt { 17 } }$$} is equal to:
(a) $$\frac { \sqrt { 29 } }{ 3 }$$
(b) $$\frac{29}{3}$$
(c) $$\frac { \sqrt { 3 } }{ 29 }$$
(d) $$\frac{3}{29}$$
(d) $$\frac{3}{29}$$

Question 2.
Fill in the blanks:

1. tan-1(1) + tan-1(2) + tan-1 (3) = …………………………..
2. tan-1(2) – tan-1 (1) = …………………………..
3. cot-1 3 + cosec-1 $$\sqrt { 5 }$$ = ……………………………
4. sin(sin-1 x + 2 cos-1 x) = ……………………………….
5. If sin-1($$\frac { 2a }{ 1+a^{ 2 } }$$) + sin-1 ($$\frac { 2b }{ 1+b^{ 2 } }$$) = 2 tan-1 x, then x = ……………………….
6. If tan-1 $$\frac{1-x}{1+x}$$ = $$\frac{1}{2}$$ tan-1 x ( When x > 0), then x = ………………………..
7. tan-1$$\frac{a-b}{1+ab}$$ + tan-1$$\frac{b-c}{1+bc}$$ + tan-1c = ………………………….

1. π
2. tan-1$$\frac{1}{3}$$
3. $$\frac{π}{4}$$
4. x
5. $$\frac{a+b}{1-ab}$$
6. $$\frac { 1 }{ \sqrt { 3 } }$$
7. tan-1 (a)

Question 3.
Write True/False:

1. tan-1x + tan-1 y = tan-1$$\frac{x+y}{1-xy}$$
2. cos-1(-x) = – cos-1 x
3. sin-1(3x – 4x3) = sin-1 $$\frac{x}{3}$$
4. cos-1 ($$\frac { 1-x^{ 2 } }{ 1+x^{ 2 } }$$) = 2 tan-1x
5. sin-1x – sin-1[xy – $$\sqrt { 1-x^{ 2 } }$$ $$\sqrt { 1-y^{ 2 } }$$]

1. True
2. False
3. False
4. True
5. False

Question 4.
Match the column:

1. (b)
2. (e)
3. (f)
4. (a)
5. (c)
6. (d)

Question 5.
Write the answer in one word/sentence:

1. Find the value of tan-1$$\frac{1}{2}$$ + tan-1$$\frac{3}{2}$$
2. Find the value of tan-1$$\frac { x }{ \sqrt { 1-x^{ 2 } } }$$
3. Find the value of sin ( 2 sin-1$$\frac{3}{5}$$)
4. Solve the equation: sin-1$$\frac{x}{5}$$ + cosec -1$$\frac{5}{4}$$ = $$\frac { \pi }{ 2 }$$
5. Write the principal value of cos-1 (cos $$\frac { 7\pi }{ 6 }$$)
6. If cos-1 ($$\frac{1}{x}$$) = θ, then find the value of tan θ

1. tan-1 8
2. sin-1x
3. $$\frac{24}{25}$$
4. x = 3
5. $$\frac { 5\pi }{ 6 }$$
6. $$\sqrt { x^{ 2 }-1 }$$

Inverse Trigonometric Functions Very Short Answer Type Questions

Question 1.
Find the principle value of the following?

1. tan-1[sin(- $$\frac { \pi }{ 2 }$$) ] (CBSE 2014)
2. cot [ $$\frac { \pi }{ 2 }$$ – 2 cot-1 $$\sqrt{3}$$ ] (CBSE 2014)
3. tan-1(- $$\sqrt{3}$$ )
4. sec-1 ( $$\frac{-2}{3}$$ $$\sqrt{3}$$ ) (NCERT)
5. cosec-1(2) (NCERT)

Solution:
1. Let, tan-1[ sin(- $$\frac { \pi }{ 2 }$$ ) = θ
⇒ tan-1 [-sin $$\frac { \pi }{ 2 }$$ ] = θ
⇒ tan-1 (-1) = θ
⇒ tan θ = 1
⇒ tan θ = – tan $$\frac { \pi }{ 4 }$$
⇒ tan θ = tan ( $$\frac { -\pi }{ 4 }$$ )
θ = $$\frac { -\pi }{ 4 }$$
∴ The principle value is $$\frac { -\pi }{ 4 }$$

2. cot($$\frac { -\pi }{ 2 }$$ – cot-1 $$\sqrt{3}$$)
Let cot-1 $$\sqrt{3}$$ = θ
⇒ cot θ = $$\sqrt{3}$$
⇒ cot θ = cot $$\frac { \pi }{ 6 }$$
∴ θ = $$\frac { \pi }{ 6 }$$
∴ cot ( $$\frac { \pi }{ 2 }$$ – cot-1$$\sqrt{3}$$ ) = cot ($$\frac { \pi }{ 2 }$$ – 2 × $$\frac { \pi }{ 6 }$$)
= cot ( $$\frac { \pi }{ 2 }$$ – $$\frac { \pi }{ 3 }$$ )
= cot $$\frac { \pi }{ 6 }$$
= $$\sqrt{3}$$
∴ The principal value is $$\sqrt{3}$$

3. Let tan(- $$\sqrt{3}$$) = θ
⇒ tan θ = – $$\sqrt{3}$$
⇒ tan θ = – tan ($$\frac { \pi }{ 3 }$$)
⇒ tan θ = tan (- $$\frac { \pi }{ 3 }$$)
⇒ θ = – $$\frac { \pi }{ 3 }$$
Hence the principle value is – $$\frac { \pi }{ 3 }$$

4. Let sec-1( $$\frac{-2}{3}$$ $$\sqrt{3}$$ ) = θ
⇒ sec-1( $$\frac { -2 }{ \sqrt { 3 } }$$ ) = θ
⇒ sec θ = $$\frac { -2 }{ \sqrt { 3 } }$$
⇒ sec θ = – sec ( $$\frac { \pi }{ 6 }$$ )
⇒ sec θ = sec (π – $$\frac { \pi }{ 6 }$$ )
⇒ sec θ = sec $$\frac { 5\pi }{ 6 }$$
∴θ = $$\frac { 5\pi }{ 6 }$$
The principle value is $$\frac { 5\pi }{ 6 }$$

5. Let cosec-1(2) = θ
⇒ cosec θ = 2
⇒ cosec θ = cosec $$\frac { \pi }{ 6 }$$ θ ∈ [- $$\frac { \pi }{ 2 }$$, $$\frac { \pi }{ 2 }$$ ]
The principle value is $$\frac { \pi }{ 6 }$$.

Question 2.
Prove the following:

1. 2 cos-1($$\frac{4}{5}$$) = cos-1( $$\frac{7}{25}$$ )
2. 2 sin-1( $$\frac{5}{13}$$ ) = sin-1( $$\frac{120}{169}$$ )
3. 2 sin-1( $$\frac{3}{5}$$ ) = sin-1( $$\frac{24}{25}$$ )

Solution:
1. 2 cos-1( $$\frac{4}{5}$$ ) = cos-1( $$\frac{7}{25}$$ )
Formula 2 cos-1 x = cos-1(2x2 – 1)
∴ L.H.S = 2 cos-1( $$\frac{4}{5}$$ )
= cos-1 (2 $$\frac{16}{25}$$ – 1)
= cos-1 ( $$\frac{32}{25}$$ – 1)
= cos-1( $$\frac{32-25}{25}$$ )
= cos-1$$\frac{7}{25}$$
= R.H.S.

2. 2 sin-1$$\frac{3}{5}$$ = sin-1$$\frac{24}{25}$$
Formula 2 sin-1(x) = sin-1(2x$$\sqrt { 1-x^{ 2 } }$$)
∴ 2 sin-1 $$\frac{3}{5}$$ = sin-1[2. $$\frac{3}{5}$$ $$\sqrt { 1-\frac { 9 }{ 25 } }$$]
= sin-1[ $$\frac{6}{5}$$ $$\sqrt { \frac { 16 }{ 25 } }$$ ]
= sin-1[ $$\frac{6}{5}$$ . $$\frac{4}{5}$$ ]
= sin-1[ $$\frac{24}{25}$$ ]
= R.H.S. Proved.

3. 2 sin-1($$\frac{5}{13}$$) = sin-1$$\frac{120}{169}$$
Solve like Q.2(b)

Question 3.
Find the value of tan-1{2 cos(2 sin-1$$\frac{1}{2}$$} (CBSE 2013, NCERT)
Solution:
tan-1[2 cos(2 sin-1$$\frac{1}{2}$$) ]
= tan-1[ 2 cos (2 sin-1 sin $$\frac { \pi }{ 6 }$$) ]
= tan-1[ 2 cos (2. $$\frac { \pi }{ 6 }$$) ]
= tan-1[ 2 cos $$\frac { \pi }{ 3 }$$ ]
= tan-1 [ 2. $$\frac{1}{2}$$ ]
= tan-1 1
= $$\frac { \pi }{ 4 }$$.

Question 4.
Find the value of sin [ $$\frac { \pi }{ 3 }$$ – sin-1( $$\frac{-1}{2}$$ ) ]? [CBSE 2008, 2013]
Solution:
sin[ $$\frac { \pi }{ 3 }$$ – sin-1($$\frac{-1}{2}$$ ) ] = sin-1 [ $$\frac { \pi }{ 3 }$$ – ( – sin-1$$\frac{1}{2}$$ ) ]
= sin-1 [ $$\frac { \pi }{ 3 }$$ + sin-1 sin$$\frac { \pi }{ 6 }$$ ]
= sin-1 [ $$\frac { \pi }{ 3 }$$ + $$\frac { \pi }{ 6 }$$ ]
= sin-1 ( $$\frac { \pi }{ 2 }$$ )
= 1.

Question 5.
Prove that:
2 tan-1$$\frac{1}{5}$$ = tan-1 ( $$\frac{5}{12}$$ )
Solution:
2 tan-1( $$\frac{1}{5}$$ ) = tan-1 ( $$\frac{5}{12}$$ )
L.H.S. = 2 tan-1 ( $$\frac{1}{5}$$ )

= tan-1 [ $$\frac{2×25}{5×24}$$ ]
= tan-1 [ $$\frac{5}{12}$$ ]
= R.H.S. Proved.

Question 6.
Find the value of tan [ 2 tan-1 $$\frac{1}{5}$$ – $$\frac { \pi }{ 4 }$$ ]?
solution:

= tan tan-1( $$\frac{-7}{17}$$ )
= $$\frac{-7}{17}$$.

Question 7.
Prove that: 3 sin-1 x = sin-1 (3x – 4x3)? (NCERT, CBSE 2018)
Solution:
Let sin-1 x = θ
⇒ x = sin θ
We know that sin 3θ = 3 sinθ – 4 sin3 θ
= 3x – 4x3
⇒ 3θ = sin-1 ( 3x – 4x3)
⇒ 3.sin-1 x = sin-1(3x – 4x3). proved.

Question 8.
Prove that: 3 cos-1 x = cos-1 (4x3 – 3x)? (NCERT)
Solution:
Let cos-1 x = cos θ
⇒ x = cos θ
We know that cos 3θ = 4 cos3θ – 3 cos θ
= 4x3 – 3x
⇒ 3θ = cos-1 (4x3 – 3x)
⇒ 3 cos-1x = cos-1 (4x3 – 3x). Proved.

Question 9.
Prove the following:

1. tan-1 $$\frac{1}{2}$$ + tan-1 $$\frac{1}{3}$$ = $$\frac { \pi }{ 4 }$$
2. cos-1 $$\frac{12}{13}$$ = tan-1 $$\frac{5}{12}$$
3. cos-1 $$\frac{3}{5}$$ = sin-1 $$\frac{4}{5}$$

Solution:
1. tan-1 $$\frac{1}{2}$$ + tan-1 $$\frac{1}{3}$$ = $$\frac { \pi }{ 4 }$$
L.H.S = tan-1 $$\frac{1}{2}$$ + tan-1 $$\frac{1}{3}$$

∴ A = tan-1 $$\frac{5}{12}$$
$$\frac { \pi }{ 4 }$$ = R.H.S. Proved

2. cos-1 $$\frac{12}{13}$$ = tan-1 $$\frac{5}{12}$$
Let cos-1 $$\frac{12}{13}$$ = A
$$\frac{12}{13}$$ = cos A
sin A = $$\sqrt { 1-cos^{ 2 }A }$$ = $$\sqrt { 1-\frac { 144 }{ 169 } }$$
= $$\sqrt { \frac { 25 }{ 169 } }$$ = $$\frac{5}{13}$$
tan A = $$\frac { sinA }{ cosA }$$ = $$\frac { 5/13 }{ 12/13 }$$ = $$\frac{5}{12}$$
A = tan-1 $$\frac{5}{12}$$
From eqns. (1) and (2), L.H.S = R.H.S. Proved.

3. cos-1 $$\frac{3}{5}$$ = sin-1 $$\frac{4}{5}$$
Let cos-1 $$\frac{3}{5}$$ = A ……………… (1)
⇒ cos A = $$\frac{3}{5}$$
⇒ sin A = $$\sqrt { 1-cos^{ 2 }A }$$
= $$\sqrt { \frac { 16 }{ 25 } }$$ = $$\frac{4}{5}$$ ……………… (2)
A = sin-1 $$\frac{4}{5}$$.
From eqns. (1) and (2), L.H.S = R.H.S. Proved.

Question 10.
Prove that:

1. sec-1 x + cosec-1 x = $$\frac { \pi }{ 2 }$$
2. sin-1x + cos-1x = $$\frac { \pi }{ 2 }$$
3. tan-1x + cot-1x = $$\frac { \pi }{ 2 }$$

Solution:
1. sec-1 x + cosec-1x = $$\frac { \pi }{ 2 }$$
Let sec -1 x = θ
∴x = sec θ
⇒ x = cosec ( $$\frac { \pi }{ 2 }$$ – θ)
⇒ cosec -1 x = $$\frac { \pi }{ 2 }$$ – θ. Proved.

2. sin-1 x + cos-1 x = $$\frac { \pi }{ 2 }$$
Let sin-1 x = θ ……………………. (1)
⇒ x = sin θ
⇒ x = cos ( $$\frac { \pi }{ 2 }$$ – θ)
⇒ cos -1 x = $$\frac { \pi }{ 2 }$$ – θ ………………. (2)
sin -1 x + cos -1 x = θ + $$\frac { \pi }{ 2 }$$ – θ
⇒ sin -1 x + cos-1 x = $$\frac { \pi }{ 2 }$$ Proved.

3. tan -1 x + cot-1 x = $$\frac { \pi }{ 2 }$$
Let tan -1 x = θ
⇒ x = tan θ
⇒ x = cot ( $$\frac { \pi }{ 2 }$$ – θ)
⇒ cot -1 x = $$\frac { \pi }{ 2 }$$ – θ
tan -1 x + cot -1 x = $$\frac { \pi }{ 2 }$$. Proved.

Question 11.
Prove the following:

1. tan-1 5 – tan-1 3 = tan-1 $$\frac{1}{8}$$
2. tan-1 3 – tan-1 2 = tan-1 $$\frac{1}{7}$$
3. tan-1 7 – tan-1 5 = tan-1 3 = tan-1 $$\frac{1}{18}$$

Solution:
1. tan-1 5 – tan-1 3 = tan-1 $$\frac{1}{8}$$
L.H.S. = tan-1 5 – tan-1 3
= tan-1 $$\frac{5-3}{1+5.3}$$ = tan-1 $$\frac{2}{16}$$ = tan-1 $$\frac{1}{8}$$
= R.H.S. Proved.

2. Solve like Q.No. 11 (A).

3. Solve like Q.No. 11(A).

Question 12.
Prove that:

1. tan-1 $$\frac{4}{7}$$ – tan-1 $$\frac{1}{5}$$ = tan-1 $$\frac{1}{3}$$
2. tan-1 $$\frac{1}{2}$$ – tan-1 $$\frac{2}{9}$$ = tan-1 $$\frac{1}{4}$$
3. tan-1 $$\frac{1}{7}$$ + tan-1 $$\frac{1}{8}$$ = tan-1 $$\frac{3}{11}$$

Solution:
1. tan-1 $$\frac{4}{7}$$ – tan-1 $$\frac{1}{5}$$ = tan-1 $$\frac{1}{3}$$

2. Solve like Q.No. 12 (A).

3. Solve like Q.No. 12 (A).

Question 13.
tan-1 1 + tan-1 2 + tan-1 3 = π?
Solution:
L.H.S. = tan-1 1 + (tan-1 2 + tan-1 3)
= tan-1 (1) + π + tan-1 ( $$\frac{2+3}{1-2×3}$$ ),
[∵ tan-1 x + tan-1 y = π + tan-1 $$\frac{x+y}{1-xy}$$, if x > 0, y > 0, xy > 1 Here xy = 6 > 1]
= tan-1 )1) + π + tan-1( $$\frac{5}{1-6}$$ )
= tan-1 (1) + π + tan-1 (-1)
= tan-1 (1) + π – tan-1 (1), [∵tan-1 (-x) = – tan-1 x]
= π = R.H.S. Proved.

Question 14.
(A) If tan-1 ( $$\frac{1}{2}$$ ) + tan-1 ( $$\frac{1}{k}$$ ) = $$\frac { \pi }{ 4 }$$ then find the value of k?
Solution:
tan-1 ( $$\frac{1}{2}$$ ) + tan-1 ( $$\frac{1}{k}$$ ) = $$\frac { \pi }{ 4 }$$

⇒ $$\frac{k+2}{2k – 1}$$ = 1
⇒ k + 2 = 2k – 1
⇒2 + 1 = 2k – k
⇒ k = 3.

(B) If tan -1 ( $$\frac{1}{2}$$ ) + tan-1 ( $$\frac{1}{k}$$ ) = $$\frac { \pi }{ 4 }$$ then find the value of k?
Solution:
Solve like Q.No. 14 (A).

(C) If tan -1 ( $$\frac{4}{5}$$ ) + tan-1 ( $$\frac{1}{k}$$ ) = $$\frac { \pi }{ 4 }$$ then find the value of k?
Solution:
Solve like Q.No. 14 (A).

Question 15.
Prove that:
tan -1 $$\sqrt { x }$$ = $$\frac{1}{2}$$ cos-1 ( $$\frac{1-x}{1+x}$$ )?
Solution:
R.H.S = $$\frac{1}{2}$$ cos -1 ( $$\frac{1-x}{1+x}$$ )
Let $$\sqrt { x }$$ = tan θ
⇒ x = tan2 θ
⇒ $$\frac{1-x}{1+x}$$ = $$\frac { 1-tan^{ 2 }\theta }{ 1+tan^{ 2 }\theta }$$ = cos 2θ
∴ R.H.S. = $$\frac{1}{2}$$ cos -1(cos 2θ)
= $$\frac{1}{2}$$ × 2θ = θ
= tan-1 ( $$\sqrt { x }$$ ) [∵$$\sqrt { x }$$ = tan θ ⇒ tan-1 ( $$\sqrt { x }$$ ) = θ]
= L.H.S. Proved.

Question 16.
Prove that:
sin (cos-1 x ) = cos (sin-1 x)?
Solution:
L.H.S. = sin(cos-1 x)
= sin [ $$\frac { \pi }{ 2 }$$ – sin -1 x],
[∵ sin-1 x + cos-1x = $$\frac { \pi }{ 2 }$$ , cos-1x = $$\frac { \pi }{ 2 }$$ – sin-1 x]
= cos (sin-1 x), [ ∵sin (90° – θ) = cos θ ]
= R.H.S. Proved.

Question 17.
(A) Prove that:
tan-1 ( $$\frac{b-c}{1+bc}$$ ) + tan-1 $$\frac{b-c}{1+bc}$$ + tan-1 c = tan-1 b?
Solution:
L.H.S = tan-1 ( $$\frac{b-c}{1+bc}$$ ) + tan-1 $$\frac{c-a}{1+ca}$$ + tan-1 a
= (tan-1 a – tan-1 b ) + (tan-1 b – tan-1 c) + tan-1 c
= tan-1 b – tan-1 c + tan-1 c – tan-1 a + tan-1 a
= tan-1 b = R.H.S. Proved.

(B) Prove that:
tan-1 ( $$\frac{a-b}{1+ab}$$ ) + tan-1 $$\frac{b-c}{1+bc}$$ + tan-1 c = tan-1 a?
Solution:
L.H.S = tan-1 ( $$\frac{a-b}{1+ab}$$ ) + tan-1 $$\frac{b-c}{1+bc}$$ + tan-1 c
= (tan-1 a – tan-1 b) + (tan-1 b – tan-1c) + tan-1 c
= tan-1 a = R.H.S. Proved.

(C) Prove that:
tan-1 $$\frac{1}{7}$$ + tan-1 $$\frac{1}{13}$$ = tan-1 $$\frac{2}{9}$$?
Solution:
tan-1 $$\frac{1}{7}$$ + tan-1 $$\frac{1}{13}$$ = tan-1 $$\frac{2}{9}$$
L.H.S = tan-1 $$\frac{1}{7}$$ + tan-1 $$\frac{1}{13}$$

= tan-1 $$\frac{20}{91}$$ × $$\frac{91}{90}$$ = tan-1 $$\frac{20}{90}$$ = tan-1 $$\frac{2}{9}$$
= R.H.S. Proved.

Question 18.
Solve the equation:
sin-1 $$\frac { 2a }{ 1+a^{ 2 } }$$ + sin-1 $$\frac { 2b }{ 1+a^{ 2 } }$$ = 2 tan-1x?
Solution:
sin-1 $$\frac { 2a }{ 1+a^{ 2 } }$$ + sin-1 $$\frac { 2b }{ 1+a^{ 2 } }$$ = 2 tan-1x, (given)
⇒ 2 tan-1 a + 2 tan-1 b = 2 tan-1 x [∵sin-1 $$\frac { 2x }{ 1+a^{ 2 } }$$ = 2 tan-1 x]
⇒ tan-1 a + tan-1 b = tan-1 x
⇒ tan-1 $$\frac{a+b}{1-ab}$$ = tan-1 x
∴ x = $$\frac{a+b}{1-ab}$$.

Question 19.
solve the equation:
cos-1 ( $$\frac { 1-a^{ 2 } }{ 1+a^{ 2 } }$$ ) – cos-1 ( $$\frac { 1-b^{ 2 } }{ 1+b^{ 2 } }$$ ) = 2 tan-1x?
Solution:
cos-1 ( $$\frac { 1-a^{ 2 } }{ 1+a^{ 2 } }$$ ) – cos-1 ( $$\frac { 1-b^{ 2 } }{ 1+b^{ 2 } }$$ ) = 2 tan-1x, (given)
⇒ 2 tan-1 a – 2 tan-1 b = 2 tan-1 x
⇒ tan-1a – tan-1 b = tan-1 x
⇒ tan-1 $$\frac{a-b}{1+ab}$$ = tan-1 x
∴ x = $$\frac{a-b}{1+ab}$$.

Question 20.
(A) Prove the following:
2 tan-1 $$\frac{1}{4}$$ = tan-1 $$\frac{8}{15}$$?
Solution:
∵ 2 tan-1 x = tan-1 ( $$\frac { 2x }{ 1-x^{ 2 } }$$ )

L.H.S = tan-1 $$\frac{16}{2.5}$$
= tan-1 $$\frac{8}{15}$$ = R.H.S. Proved.

(B) Prove the following:
2 tan-1 $$\frac{1}{2}$$ = tan-1 $$\frac{4}{3}$$?
Solution:
We know that 2 tan-1 x = tan-1 ( $$\frac { 2x }{ 1-x^{ 2 } }$$ )

= tan-1 ( $$\frac{4}{3}$$ )
= R.H.S. Proved.

Question 21.
Write in simplest form
tan-1 $$\sqrt { \frac { 1-cosx }{ 1+cosx } }$$?
Solution:

Question 22.
Write in simplest form:
cos-1 $$\sqrt { \frac { 1 }{ 2 } (1+cosx) }$$?
Solution:
cos-1 $$\sqrt { \frac { 1 }{ 2 } (1+cosx) }$$ = cos-1 $$\sqrt { \frac { 1 }{ 2 } .2cos^{ 2 }\frac { x }{ 2 } }$$
= cos-1 (cos $$\frac{x}{2}$$ ) = $$\frac{x}{2}$$.

Question 23.
If tan-1 a + tan-1 b + tan-1 c = $$\frac { \pi }{ 2 }$$ then prove that ab + bc + ca = 1?
Solution:
tan-1 a + tan-1 b + tan-1 c = $$\frac { \pi }{ 2 }$$ , given
⇒ tan-1 a + tan-1 b + tan -1 c = tan-1 a + cot-1 a, [∵tan-1 a + cot -1 a = $$\frac { \pi }{ 2 }$$ ]
⇒ tan-1 b + tan-1c = cot-1 a
⇒ tan-1 ( $$\frac{b+c}{1+bc}$$ ) = $$\frac{1}{a}$$
⇒ ab + ca = 1 – bc
⇒ ab + bc + ca = 1. Proved.

Question 24.
Prove that:
tan-1 $$\frac{2}{11}$$ + cot-1 $$\frac{24}{7}$$ = tan-1 $$\frac{1}{2}$$?
Solution:
L.H.S. = tan-1 $$\frac{2}{11}$$ + cot-1 $$\frac{7}{24}$$
= tan-1 $$\frac{2}{11}$$ + tan-1 $$\frac{7}{24}$$

tan-1 = $$\frac{125}{250}$$ = tan-1 $$\frac{1}{2}$$ = R.H.S.

Question 25.
Prove that:
cos-1 x = 2 cos-1 $$\sqrt { \frac { 1+x }{ 2 } }$$?
Solution:
R.H.S. = 2 cos-1 $$\sqrt { \frac { 1+x }{ 2 } }$$
= 2 cos-1 $$\sqrt { \frac { 1+cos\theta }{ 2 } }$$

= cos-1 x
= L.H.S. Proved.

Question 26.
Prove that:
cos-1 x = 2 tan-1$$\sqrt { \frac { 1-x }{ 1+x } }$$?
Solution:
R.H.S. = 2 tan-1 $$\sqrt { \frac { 1-x }{ 1+x } }$$
= 2 tan-1 $$\sqrt { \frac { 1-cos\theta }{ 1+cos\theta } }$$ (putting x = cos θ)

= 2. $$\frac { \theta }{ 2 }$$ = θ = cos-1 x = L.H.S. Proved.

Question 27.
tan-1 ( $$\frac{a}{b}$$ ) – tan-1 ( $$\frac{a-b}{a+b}$$ ) = $$\frac { \pi }{ 4 }$$?
Solution:
tan-1 ( $$\frac{a}{b}$$ ) – tan-1 ( $$\frac{a-b}{a+b}$$ ) = $$\frac { \pi }{ 4 }$$

⇒ 1 = tan$$\frac { \pi }{ 4 }$$
or tan$$\frac { \pi }{ 4 }$$ = 1. Proved.

Inverse Trigonometric Functions Long Answer Type Questions – I

Question 1.
(A) Prove that:
sin-1 $$\frac { 1 }{ \sqrt { 5 } }$$ + sin-1 $$\frac { 1 }{ \sqrt { 10 } }$$ = $$\frac { \pi }{ 4 }$$?
Solution:
Let sin-1 $$\frac { 1 }{ \sqrt { 5 } }$$ = A, sin-1 $$\frac { 1 }{ \sqrt { 10 } }$$ = B
∴ A + B = $$\frac { \pi }{ 4 }$$
⇒ sin (A + B) = sin$$\frac { \pi }{ 4 }$$
⇒ sin (A + B) = sin$$\frac { \pi }{ 4 }$$
⇒ sin A cos B + cos A sin B = $$\frac { 1 }{ \sqrt { 2 } }$$
L.H.S = sin A $$\sqrt { 1-sin^{ 2 }B }$$ + $$\sqrt { 1-sin^{ 2 }A }$$. sin B
= $$\frac { 1 }{ \sqrt { 5 } }$$. $$\sqrt { 1-\frac { 1 }{ 10 } }$$ + $$\sqrt { 1-\frac { 1 }{ 5 } }$$. $$\frac { 1 }{ \sqrt { 10 } }$$
= $$\frac { 3 }{ \sqrt { 5.\sqrt { 10 } } }$$ + $$\frac { 2 }{ \sqrt { 5.\sqrt { 10 } } }$$
= $$\frac { 5 }{ \sqrt { 5.\sqrt { 10 } } }$$ = $$\sqrt { \frac { 5 }{ 10 } }$$ = $$\frac { 1 }{ \sqrt { 2 } }$$
= R.H.S. Proved.

(B) Solve the following equation:
sin-1 x + sin-1 (1 – x) = sin-1 $$\sqrt { 1-x^{ 2 } }$$?
Solution:
Let sin-1 x = α ∴ x = sin α
Here α + sin-1 ( 1 – sin α) = sin-1 $$\sqrt { 1-sin^{ 2 }\alpha }$$
⇒ α + sin-1 ( 1 – sin α) = sin-1 cos α
⇒ α + sin-1 ( 1 – sin α) = sin-1. sin ( $$\frac { \pi }{ 2 }$$ – α)
⇒ α + sin-1 ( 1 – sin α) = $$\frac { \pi }{ 2 }$$ – α
⇒ sin-1 ( 1 – sin α) = $$\frac { \pi }{ 2 }$$ – 2α
⇒ 1 – sin α = sin ( $$\frac { \pi }{ 2 }$$ – 2α)
⇒ 1 – sin α = cos 2α
⇒ 1 – cos 2α = sin α
⇒ 2 sin2α = sin α
⇒ sin α = $$\frac{1}{2}$$ ∴α = $$\frac { \pi }{ 6 }$$
or x = $$\frac { \pi }{ 6 }$$

Question 2.

1. tan-1 $$\frac{x+1}{x}$$ – tan-1 $$\frac{1}{2x+1}$$ = $$\frac { \pi }{ 4 }$$?
2. If tan-1 x + tan-1 y + tan-1 z = π then prove that x + y + z = xyz?
3. If tan-1 x + tan-1 y + tan-1 z = $$\frac { \pi }{ 2 }$$ then prove that xy + yz + zx = 1?

Solution:
1. tan-1 $$\frac{x+1}{x}$$ – tan-1 $$\frac{1}{2x+1}$$ = $$\frac { \pi }{ 4 }$$

= R.H.S.

2. tan-1 x + tan-1 y + tan-1 z = π
⇒ tan-1 $$\frac{x+y}{1-xy}$$ + tan-1 z = π

⇒ x + y + z – xyz = 0, [∵ tan π = 0]
∴ x + y + z = xyz. Proved.

3. Solve like Q.2(B), take tan $$\frac { \pi }{ 4 }$$ = ∞ = $$\frac{1}{0}$$

Question 3.
Write in simplest form:
tan-1 [ $$\frac { \sqrt { 1+x^{ 2 }-1 } }{ x }$$ ]?
Solution:
tan-1 [ $$\frac { \sqrt { 1+x^{ 2 }-1 } }{ x }$$ ]
Let x = tan θ

Question 4.
(A) Prove the following:
$$\frac{1}{2}$$ sin-1 x = cot-1 [ $$\frac { \sqrt { 1+x^{ 2 }-1 } }{ x }$$ ]?
Solution:
R.H.S = cot-1 [ $$\frac { \sqrt { 1+x^{ 2 }-1 } }{ x }$$ ]

= L.H.S. Proved.

(B) Prove that:
$$\frac{1}{2}$$ cot-1 x = cot-1 ( $$\sqrt { 1+x^{ 2 }+x }$$ )?
Solution:
$$\frac{1}{2}$$ cot-1 x = cot-1 ( $$\sqrt { 1+x^{ 2 }+x }$$ )
R.H.S = cot-1 ( $$\sqrt { 1+x^{ 2 }+x }$$ )
Let x = cos θ
R.H.S = cot-1 ( $$\sqrt { 1+cot^{ 2 }\theta }$$ + cot θ )
= cot-1 ( $$\sqrt { cosec^{ 2 }\theta }$$ + cot θ )
= cot-1 (cosec θ + cot θ)

= $$\frac{1}{2}$$ cot-1 x
= L.H.S. Proved.

Question 5.
Solve the following equation:
tan-1 (x + 1) + tan-1 (x – 1) = tan-1 ( $$\frac{6}{17}$$ )?
Solution:
Given: tan-1 (x + 1) + tan-1 (x – 1) = tan-1 ( $$\frac{6}{17}$$ )
⇒ tan-1 (x + 1) + tan-1 (x – 1) = tan-1 ( $$\frac{6}{17}$$ )

⇒ 17 x = 6 – 3x2
⇒ 3x2 + 17x – 6 = 0
⇒ 3x2 + 18x – x – 6 = 0
⇒ 3x (x + 6) – 1 (x + 6) = 0
⇒ (x+6) (3x – 1) = 0
∴ x = – 6, x = $$\frac{1}{3}$$

Question 6.
Prove that cos-1 $$\frac{3}{5}$$ + cos-1 $$\frac{4}{5}$$ = $$\frac { \pi }{ 2 }$$?
Solution:
L.H.S = cos-1 $$\frac{3}{5}$$ + cos-1 $$\frac{4}{5}$$,

= $$\frac { \pi }{ 2 }$$ = R.H.S Proved.

Question 7.
If cos-1 x + cos-1 y + cos-1 z = π then prove that:
x2 + y2 + z2 + 2xyz = 1?
Solution:
Given: cos-1 x + cos-1 z = π
⇒ cos-1 x + cos-1 y = π- cos-1 z

Squaring on both sides
x2y2 + z2 + 2xyz = (1 – x2) (1 – y2)
⇒ x2y2 + z2 + 2xyz = 1 – y2 – x2 + x2y2
⇒ z2 + 2xyz = 1 – y2 – x2
⇒ x2 + y2 + z2 + 2xyz = 1. Proved.

Question 8.
If sin-1 $$\frac { 2a }{ 1+a^{ 2 } }$$ – cos-1 $$\frac { 1-b^{ 2 } }{ 1+b^{ 2 } }$$ = tan-1 $$\frac { 2x }{ 1-x^{ 2 } }$$ then prove that:
x = $$\frac{a-b}{1+ab}$$?
Solution:

⇒ sin-1 (sin 2θ) – cos-1 (cos 2ϕ) = tan-1 (tan 2Ψ)
⇒ 2θ – 2ϕ = 2Ψ
⇒ θ – ϕ = Ψ
⇒ tan-1 a – tan-1 b = tan-1 x
⇒ tan-1 ( $$\frac{a-b}{1+ab}$$ ) = tan-1 x
⇒ x = $$\frac{a-b}{1+ab}$$. Proved.

Question 9.
Prove the following

Solution:
Let x = cos θ, then θ = cos-1 x.

$$\frac { \pi }{ 4 }$$ – $$\frac{1}{2}$$ cos-1 x.

Question 10.
Write in simplest form:
tan-1 ( $$\frac { x }{ \sqrt { 1+x^{ 2 }-1 } }$$ )?
Solution:
tan-1 ( $$\frac { x }{ \sqrt { 1+x^{ 2 }-1 } }$$ )
Putting x = tan θ, we get

Question 11.
Prove that:
tan-1 $$\sqrt{x}$$ = $$\frac{1}{2}$$ cos-1 ( $$\frac{1-x}{1+x}$$ )? (NCERT)
Solution:
tan-1 $$\sqrt{x}$$ = $$\frac{1}{2}$$ cos-1 ( $$\frac{1-x}{1+x}$$ )
R.H.S = $$\frac{1}{2}$$ cos-1 ( $$\frac{1-x}{1+x}$$ )

Let tan-1 $$\sqrt{x}$$ = θ
$$\sqrt{x}$$ = tan θ
R.H.S. = $$\frac{1}{2}$$ cos-1(cos 2θ)
= $$\frac{1}{2}$$. 2θ
= θ
= tan-1 $$\sqrt{x}$$
= L.H.S. Proved.

Question 12.
Prove that:

Solution:

= tan-1 (tan 3θ) – tan-1 (tan 2θ)
= 3θ – 2θ
= θ
= tan-1 2x
= R.H.S. Proved.

Question 13.
Prove that:
cos-1 $$\frac{4}{5}$$ + cos-1 $$\frac{12}{13}$$ = cos-1 $$\frac{33}{65}$$?
Solution:
Let cos-1 $$\frac{4}{5}$$ = A
∴ $$\frac{4}{5}$$ = cos A
∴ sin A = $$\sqrt { 1-cos^{ 2 }A }$$ = $$\sqrt { 1-\frac { 16 }{ 25 } }$$ = $$\sqrt { \frac { 9 }{ 25 } }$$
⇒ sin A = $$\frac{3}{5}$$
Let cos-1 $$\frac{12}{13}$$ = B
⇒ $$\frac{12}{13}$$ = B
∴ sin B = $$\sqrt { 1-cos^{ 2 }B }$$ = $$\sqrt { 1-\frac { 144 }{ 169 } }$$ = $$\sqrt { \frac { 25 }{ 169 } }$$
⇒ sin B = $$\frac{5}{13}$$
A + B = cos-1 $$\frac{33}{65}$$
⇒ cos (A + B) = $$\frac{33}{65}$$
⇒ cos A.cos B – sin A.sin B = $$\frac{33}{65}$$
L.H.S. = $$\frac{4}{5}$$. $$\frac{12}{13}$$ – $$\frac{3}{5}$$. $$\frac{5}{13}$$
= $$\frac{48}{65}$$ – $$\frac{15}{65}$$
= $$\frac{33}{65}$$
= R.H.S. Proved.