NCERT Solutions for Class 12 Maths Chapter 2 Inverse Trigonometric Functions: Exercise PDFs, Weightage & Formulae

Chapter 2 Inverse Trigonometric function carries 5-6 marks in CBSE Board Exams, often short-answer or value-based questions are asked from this chapter. Inverse Trigonometric Functions have a moderate weightage in competitive exams; however, its concepts are frequently used in integral calculus problems. 

This chapter includes inverses of trigonometric functions like sine, cosine, and tangent. Check the key topics of this chapter below; 

• Inverse trigonometric functions: sin⁻¹(x), cos⁻¹(x), tan⁻¹(x), cosec⁻¹(x), sec⁻¹(x), cot⁻¹(x)

• Properties: Principle value, real value, domain, range, and co-domain of inverse trigonometric functions

Check out Dropped Topics of Inverse Trigonometric Function: 2.3 Properties of Inverse Trigonometric Functions (Except sin (sin^-1 x) = x, x ∈ [-1, 1] sin^-1 (sin x) = x, x ∈ [-π/2, π/2], Page 45 – 47 Examples 4, 7 and 8; Alternative Solution of Example 5, Page 47-48 Ques. 3, 4, 6, 12, 14, 15, Page 49-51 Examples 10, 11, 12, 13, Page 51-52 Ques. 8, 12, 17 (Miscellaneous Exercise), Page 53 Summary Points 8–13

Read more:  Most scoring topics in Mathematics for JEE Main

Inverse Trigonometric Function Important Formulae for CBSE and Competitive Exams

Basic Properties

• $\sin^{-1}(-x) = -\sin^{-1}(x), \quad x \in [-1, 1]$
  
• $\cos^{-1}(-x) = \pi - \cos^{-1}(x), \quad x \in [-1, 1]$
  
• $\tan^{-1}(-x) = -\tan^{-1}(x), \quad x \in \mathbb{R}$
  
• $\cot^{-1}(-x) = \pi - \cot^{-1}(x), \quad x \in \mathbb{R}$
  
• $\sec^{-1}(-x) = \pi - \sec^{-1}(x), \quad |x| \geq 1$
  
• $\csc^{-1}(-x) = -\csc^{-1}(x), \quad |x| \geq 1$
  

Domain and Range of ITF

Sum and Difference Formulas

• $\sin^{-1}x + \cos^{-1}x = \frac{\pi}{2}, \quad x \in [-1, 1]$
  
• $\tan^{-1}x + \cot^{-1}x = \frac{\pi}{2}, \quad x \in \mathbb{R}$
  
• $\sec^{-1}x + \csc^{-1}x = \pi, \quad |x| \geq 1$
  

We have compiled solutions for all exercises of Inverse Trigonometric Functions in one place for ease of students. Students can download the Class 12 Math Chapter 2 ITF solution PDF for free. The NCERT Solutions for Class 12 Maths Inverse Trigonometric Functions PDF is a very helpful document for all student boards as well as JEE Main, and other entrance exams. access the PDF below for free;

Class 12 Math Chapter 2 Inverse Trigonometric Function Solution: Free PDF Download

Inverse Trigonometric Function is an essential topic for building foundational trigonometric concepts. Chapter 2 Class 12 Math Inverse Trigonometric Function deals with the properties, domains, and ranges of inverse trigonometric functions. The ITF concepts also play a vital role in calculus, Matrices, Determinants, and other important physics and Math concepts and real-world applications. Students can check the exercise-wise Chapter 2 ITF math solutions below;

Chapter 2 Inverse Trigonometric Function Exercise 2.1 focuses on problems finding principal values, range, domain, and co-domain. Chapter 2 ITF Exercise 2.1 includes 14 Questions (12 Short Answers, 2 MCQs) primarily based on finding the values of inverse trigonometric functions. Students can check the complete the exercise 2.1 solutions below;

Inverse Trigonometric Function Exercise 2.1 Solutions

Find the principal values of the following:

Q1.  $si{n}^{-1}\left(-\frac{1}{2}\right)$

A.1. Let sin⁻¹ $$  $\left(\frac{-1}{2}\right)$ =y. Then, sin y=-  $\frac{-1}{2}$

We know that the range of principal value branch of sin⁻¹ is $-\frac{\pi }{2},$

and sin⁻¹ $\left(\frac{-1}{2}\right)$ =−sin⁻¹ $\frac{-1}{2}$ (sin (-x) = -sin x)

=  $\left(-\frac{\pi }{6}\right)$ = sin y (as sin $<br>\frac{\pi }{6}<br>$ $\frac{}{}$ =  $\frac{1}{2}$ )

Principal value of sin⁻¹ $\left(-\frac{1}{2}\right)$ is  $\left(-\frac{\pi }{6}\right)$

Q5.  $co{s}^{-1}\left(-\frac{1}{2}\right)$

A.5. Let cos ^-1 $\left(\frac{-1}{2}\right)$ =y Then cos y =  $\frac{-1}{2}$ = − cos $\frac{\mathrm{<br>}\pi }{3}$ $\frac{}{}$ cos  $\left(\pi -\frac{x}{3}\right)$

= cos  $\frac{3\pi -\pi }{3}$

= cos  $\frac{2\pi }{3}$

$$ (E) We know that the range of principal value

branch of cos⁻¹ is [0, $\frac{\pi }{}$] and cos $\frac{2x}{3}$ =  $\frac{-1}{2}$

Principal value of cos−1  $\left(\frac{-1}{2}\right)$ is  $\frac{2x}{3}$ .

Find the values of the following:

Q11.  $ta{n}^{-1}\left(1\right)+co{s}^{-1}-\frac{1}{2}+si{n}^{-1}-\frac{1}{2}$

A.11. tan⁻¹(1) + cos⁻¹ $\left(\frac{1}{2}\right)$ +  sin ⁻¹ $\left(\frac{-1}{2}\right)$

Q12.  $co{s}^{-1}\frac{1}{2}+2si{n}^{-1}\frac{1}{2}$

A.12. cos⁻¹ $\frac{1}{2}$ + 2Sin⁻¹ $\frac{1}{2}$ = cos⁻¹ $\left(cos\frac{\pi }{3}\right)$ + 2×Sin⁻¹ $\left(sin\frac{\pi }{6}\right)$

=  $\frac{\pi }{3}+2\times \frac{\pi }{6}$

=  $\frac{\pi }{3}+\frac{\pi }{3}$

=  $\frac{2\pi }{3}$

Q13.  $Ifsi{n}^{-1}x=y,then$

$\begin{array}{l}\left(A\right)0\le y\le \pi \left(B\right)-\frac{\pi }{2}\le y\le \frac{\pi }{2}\\ \left(C\right)0<y<\pi \left(D\right)-\frac{\pi }{2}<y<\frac{\pi }{2}\end{array}$

A.13. Given, Sin⁻¹x=y.

(E) We know that the principal value branch of Sin⁻¹ is

$[\frac{-\pi }{2},\frac{\pi }{2}]$ Hence ,  $\frac{-\pi }{2}$ ≤ y ≤  $\frac{\pi }{2}$

Option B is correct.

Q14.  $ta{n}^{-1}$ √3 $-se{c}^{-1}\left(-2\right)isequalto$

$\left(A\right)\pi \left(B\right)-\frac{\pi }{3}\left(C\right)\frac{\pi }{3}\left(D\right)\frac{2\pi }{3}$

A.14.

=tan⁻¹ $\left(\frac{tan\pi }{3}\right)$ −  $\pi$ + sec⁻¹ $\left(sec\frac{\pi }{3}\right)$

=  $\frac{\pi }{3}-\pi +\frac{\pi }{3}$

=  $\frac{\pi -3\pi +\pi }{3}$

=  $\frac{-\pi }{3}.$

Option B is correct.

Chapter 2 Inverse Trigonometric Function Exercise 2.2 consists of 21 Questions (18 Short Answers, 3 MCQs). Chapter 2 ITF contains questions related to solving ITF equations and finding the values of complex inverse trigonometric functions. Students can check the complete exercise 2.2 solutions below;

Relation and Function Exercise 2.2 Solutions

Q1.  $ta{n}^{-1}\left(\frac{x}{y}\right)-ta{n}^{-1}\frac{x-y}{x=y}isequalto$

$\left(A\right)\frac{\pi }{2}\left(B\right)\frac{\pi }{3}\left(C\right)\frac{\pi }{4}\left(D\right)\frac{3\pi }{4}$

A.1. We know that.

sin 3θ =3 sin θ 4sin³θ (identity).

(E) Let x = sinθ. Then, sin ⁻¹x=θ . We have,

Sin3(sin ⁻¹x) = 3x−4x³

3sin ⁻¹x =sin^-1 (3x−4x³)

Hence proved.

Q2.  $3co{s}^{-1}x=co{s}^{-1}\left(4x^3-3x\right),x\in \left[\frac{1}{2},1\right]$

A.2. We know that,

cos3θ= 4cos³θ - 3cosθ

Letx = cosθ Then θ = cos^-1x. We have,

Cos3(cos^-1x) = 4x³-3x

3cos^-1x = cos^-1(4x³- 3x)

Hence Proved

Q3.  $ta{n}^{-1}\frac{2}{11}+ta{n}^{-1}\frac{7}{24}=ta{n}^{-1}\frac{1}{2}$

A.3. L.H.S = tan ^-1 $\frac{2}{11}$ + tan ^-1 $\frac{7}{24}$

Using tan ^-1x+ tan ^-1y= tan ^-1 $\frac{x+y}{1-xy}$ , xy<1

L.H.S =tan ^-1 $\frac{\frac{2}{11}+\frac{7}{24}}{1-\frac{2}{11}\times \frac{7}{24}}$ = tan ^-1 $\frac{\frac{2\times 24+7\times 2}{11\times 24}}{\frac{11\times 24-7\times 2}{11\times 24}}$

tan ^-1 $\frac{48+14}{264-14}$ = tan ^-1 $\frac{125}{250}$ = tan ^-1 $\frac{1}{2}$ = R.H.S

Q4.  $2ta{n}^{-1}\frac{1}{2}+ta{n}^{-1}\frac{1}{7}=ta{n}^{-1}\frac{31}{17}$

A.4. L.H.S= 2 tan ^-1 $\frac{1}{2}$ + tan ^-1 $\frac{1}{7}$

(E) Using2 tan ^-1x= tan ^-1 $\frac{2a}{1-x^2}$ we can write.

L.H.S = tan ^-1 $\frac{2\times \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}{1-{\left(\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\right)}^2}$ + tan ^-1 $\frac{1}{7}$

= tan ^-1 $\frac{1}{1\raisebox{1ex}{$-1$}\!\left/ \!\raisebox{-1ex}{$4$}\right.}$ + tan ^-1 $\frac{1}{7}$

= tan ^-1 $\frac{1}{4\raisebox{1ex}{$-1$}\!\left/ \!\raisebox{-1ex}{$4$}\right.}$ + tan ^-1 $\frac{1}{7}$ = tan ^-1 $\frac{4}{3}$ + tan ^-1 $\frac{1}{7}$

= tan ^-1 $\frac{\frac{4}{3}+\frac{1}{7}}{1-\frac{4}{3}\times \frac{1}{7}}$ { Ø tan -1 x + tan -1 y = tan - 1 $\frac{x+y}{1-xy}$ }

= tan ^-1 $\frac{\frac{7\times 4+1\times 3}{3\times 7}}{\frac{3\times 7-4\times 1}{3\times 7}}$

= tan -1  $\frac{28+3}{21-4}$ = tan ^-1 $\frac{31}{17}$ = R H S

Write the following functions in the simplest form:

= tan ^-1 $\frac{sec-\theta }{tan\theta }$

tan ^-1 $\frac{\frac{1}{cos\theta }-1}{\frac{sin\theta }{cos\theta }}$ = tan ^-1 $\frac{\frac{1-cos\theta }{sin\theta }}{\frac{sin\theta }{cos\theta }}$  $=ta{n}^{+}\frac{1-cos\theta }{sin\theta }=ta{n}^{-1}\frac{2si{n}^2\raisebox{1ex}{$\theta $}\!\left/ \!\raisebox{-1ex}{$2$}\right.}{2sin\raisebox{1ex}{$\theta $}\!\left/ \!\raisebox{-1ex}{$2$}\right.cos\raisebox{1ex}{$\theta $}\!\left/ \!\raisebox{-1ex}{$2$}\right.}$

$=la{n}^{-1}\frac{sin\raisebox{1ex}{$\theta $}\!\left/ \!\raisebox{-1ex}{$2$}\right.}{cos\raisebox{1ex}{$\theta $}\!\left/ \!\raisebox{-1ex}{$2$}\right.}$ { Ø cos 2Ø = 1 - sin²ØÞ 2 sin2 Ø = 1 - cos 2Ø => 2 sin2 $\raisebox{1ex}{$\theta $}\!\left/ \!\raisebox{-1ex}{$2$}\right.$ = 1 cos2  $\raisebox{1ex}{$\theta $}\!\left/ \!\raisebox{-1ex}{$2$}\right.$ = 1- cosØ }

$=ta{n}^{-1}\left(tan\raisebox{1ex}{$\theta $}\!\left/ \!\raisebox{-1ex}{$2$}\right.\right)$ Similar

$=\raisebox{1ex}{$\theta $}\!\left/ \!\raisebox{-1ex}{$2$}\right.$ sin 2Ø = 2sinØ cosØ => sin $\frac{\theta }{2}$ = 2 sin  $\raisebox{1ex}{$\theta $}\!\left/ \!\raisebox{-1ex}{$2$}\right.$ cos  $\raisebox{1ex}{$\theta $}\!\left/ \!\raisebox{-1ex}{$2$}\right.$ => sin = 2 sin $\raisebox{1ex}{$\theta $}\!\left/ \!\raisebox{-1ex}{$2$}\right.$ cos  $\raisebox{1ex}{$\theta $}\!\left/ \!\raisebox{-1ex}{$2$}\right.$

$=\frac{ta{n}^{-1}x}{2}$

Q8.  $ta{n}^{-1}\left(\frac{cosx-sinx}{cosx+sinx}\right),0\frac{-\pi }{4}<x<\frac{3\pi }{4}$

A.8. Given tan ^-1 $\frac{cosx-sinx}{cosx+sinx}$ ,  $\frac{-\pi }{4},$ x<  $\frac{3x}{4}$

(M) Dividing numerator & denominator cos x we get,

tan ^-1 $\frac{\frac{cosx-sinx}{cosx}}{\frac{cosx+sinx}{cosx}}=ta{n}^{-1}\frac{\frac{cosx}{cosx}-\frac{sinx}{cosx}}{\frac{cosx}{cosx}+\frac{sinx}{cosx}}=ta{n}^{-1}\frac{1-tanx}{1+tanx.}$

We know that  $tan\frac{\pi }{4}=1$ = 1 so,

$=ta{n}^{-1}\frac{tan\frac{\pi }{4}-tanx}{1+tan\frac{\pi }{4}tanx}=ta{n}^{-1}\left[tan\left(\frac{\pi }{4}-x\right)\right]$  $\{\because tan(x-y)=\frac{tanx-tany}{1+tanx·tany}\}$

$=\frac{\pi }{4}-x$ .

Find the values of each of the following:

Q11.  $ta{n}^{-1}\left[2cos\left(2si{n}^1\frac{1}{2}\right)\right]$

A.11.  $ta{n}^{-1}\left[2cos\left(2si{n}^{-1}\frac{1}{2}\right)\right]=ta{n}^{-1}\left[2cos\left(2si{n}^{-1}sin\frac{\pi }{6}\right)\right]$

(E)  $=ta{n}^{-1}\left[2cos\left(2\times \frac{\pi }{6}\right)\right]$

$=ta{n}^{-1}\left[2cos\frac{\pi }{3}\right]$

$=ta{n}^{-1}2\times \frac{1}{2}$

$=ta{n}^{-1}1$

$=ta{n}^{-1}\left(tan\frac{\pi }{4}\right)$

$=\frac{\pi }{4}$

Q12.  $cot\left(ta{n}^{-1}a+co{t}^{-1}a\right)$

A.12. cot $\left(ta{n}^{-1}a+co{t}^{-1}a\right)=cot\frac{\pi }{2}\left\{\because ta{n}^{-1}x+co{t}^{-1}x=\frac{\pi }{2}\right\}$

(E)  $=0$

Q13.  $tan\frac{1}{2}\left[si{n}^{-1}\frac{2x}{1+x^2}+co{s}^{-1}\frac{1-y^2}{1+y^2}\right],\left|x\right|<1,y>0andxy<1$

A.13.  $tan\frac{1}{2}\left[si{n}^{-1}\frac{2x}{1+x^2}+co{s}^{-1}\frac{1-y^2}{1+y^2}\right],\left|x\right|<1,y>$

(M) Let x = tanØ . then tan^-1x= 0 and y = tan ω then tan^-1y = ω. we have,

tan  $\frac{1}{2}$  $\{sin{}^{-1}\frac{2tan\theta }{1+ta{n}^2\theta }+\frac{1-ta{n}^2\omega }{1+ta{n}^2\omega }\}$

$=tan\frac{1}{2}\left\{si{n}^{-1}(sin2\theta )+co{s}^{-1}(cos2\omega )\right\}$  $\left\{\therefore sin2\theta \frac{2tan\theta }{1+ta{n}^2\theta }cos2\theta =\frac{1-ta{n}^2\theta }{1+ta{n}^2\theta }tanx+y=\frac{tanx+tany}{1-tanxtany}\right\}$

$=tan\frac{1}{2}\{2\theta +2\omega \}=tan(\theta +\omega )$

$=\frac{tan\theta +tan\omega }{1-tan\theta tan\omega }$

=  $\frac{x=y}{1-xy}$

Q14.  $Ifsin\left(si{n}^1\frac{1}{5}+co{s}^{1}x\right)=1,thenfindthevalueofx$

A.14.  $sin\left(si{n}^{-1}\frac{1}{5}+co{s}^{-1}x\right)=1.$

(E)  $\Rightarrow sin\left(si{n}^{-1}\frac{1}{5}+co{s}^{-1}x\right)=sin\frac{\pi }{2}\left\{\because sin\frac{\pi }{2}=1\right\}$

$\Rightarrow si{n}^{-1}\frac{1}{5}+co{s}^{-1}x=si{n}^{-1}\left(sin\frac{\pi }{2}\right)=\frac{\pi }{2}$

$\Rightarrow co{s}^{-1}x=\frac{\pi }{2}-si{n}^{-1}\frac{1}{5}$

$\Rightarrow co{s}^{-1}x=co{s}^{-1}\frac{1}{5}$  $\left\{\because \frac{\pi }{2}=si{n}^{-1}x+co{s}^{-1}x\right\}$

$=\frac{\pi }{2}-si{n}^{-1}\frac{1}{5}=co{s}^{-1}\frac{1}{5}$  $\{\begin{array}{l}\Rightarrow \frac{\pi }{2}-si{n}^{-1}x=co{s}^{-1}x\\ x=\frac{1}{5}\\ \Rightarrow \frac{\pi }{2}-si{n}^{-1}\frac{1}{5}=co{s}^{-1}\frac{1}{5}\end{array}.$

= x =  $\frac{1}{5}.$

Q15.  $Ifta{n}^{-1}\frac{x-1}{x-2}+ta{n}^{-1}\frac{x+1}{x+2}=\frac{\pi }{4},thenfindthevalueofx$

A.15.  $ta{n}^{-1}\frac{x-1}{x-2}+ta{n}^{-1}\frac{x+1}{x+2}=\frac{\pi }{4}$

(E)Using  $ta{n}^{-1}x+ta{n}^{-1}y=ta{n}^{-1}\frac{x+y}{1-xy}.$

$\Rightarrow ta{n}^{-1}\frac{\frac{(x-1)}{(x-2)}+\left(\frac{x+1}{x+2}\right)}{1-\frac{(x-1)}{(x-2)}\times \left(\frac{x+1}{x+2}\right)}=\frac{\pi }{4}.$

$\Rightarrow \frac{\frac{(x-1)(x+2)-1(x+1)(x-2)}{(x-2)(x+2)}}{\frac{(x-2)(x+2)-(x-1)(x+1)}{(x-2)(x+2)}}=tan\frac{\pi }{4}$

$\Rightarrow \frac{(x-1)(x+2)+(x+1)(x-2)}{(x-2)(x+2)-(x-1)(x+1)}=1.$

$\Rightarrow \frac{x^2+2x-x-2+x^2-2x+x-2}{\left(x^2-4\right)-\left(x^2-1^2\right)}=1.$  $\left\{\begin{array}{cc}\hfill \because (a-b)(a+b)=\hfill & \hfill a^2-b^2\hfill \end{array}\right\}$

$\Rightarrow \frac{2x^2-4}{x^2-4-x^2+1}=1\Rightarrow \frac{2x^2-4}{-3}=1\Rightarrow 2x^2-4=-3$

Find the values of each of the expressions in Exercises 16 to 18.

Q16.  $si{n}^{-1}\left(sin\frac{2\pi }{3}\right)$

A.16.  $si{n}^{-1}\left(sin\frac{2\pi }{3}\right).$

(M) As  $\frac{2\pi }{3}\notin \left[-\frac{\pi }{2},\frac{\pi }{2}\right];$ principal value branch of sin-1 we can write,

$si{n}^{-1}\left(sin\frac{3\pi -\pi }{3}\right)=si{n}^{-1}\left(sin\frac{3\pi }{3}-\frac{\pi }{3}\right)=si{n}^{-1}\left(sin\pi -\frac{\pi }{3}\right)$

$=\frac{\pi }{3}.$

Q17.  $ta{n}^{-1}\left(tan\frac{3\pi }{4}\right)$

A.17.  $ta{n}^{-1}\left(tan\frac{3\pi }{4}\right).$

(M). As  $\frac{3\pi }{4}\notin \left(-\frac{\pi }{2},\frac{\pi }{2}\right);$ principal value branch of tan ^-1we can write,

$ta{n}^{-1}\left(tan\frac{3\pi }{4}\right)=ta{n}^{-1}tan\frac{4\pi -\pi }{4}$  $=ta{n}^{-1}\left(tan\frac{4\pi }{4}-\frac{\pi }{4}\right)$

$=ta{n}^{-1}\left(tan\pi -\frac{\pi }{4}\right)$

 quadent }

$=-ta{n}^{-1}\left(tan\frac{1}{4}\right)\left\{\because ta{n}^{-1}(-x)=-ta{n}^{-1}x\right\}$

=  $-\frac{\pi }{4}$

Q18.  $tan\left(si{n}^{-1}\frac{3}{5}+co{t^{-}}^1\frac{3}{2}\right)$

A.18.  $tan\left(si{n}^{-1}\frac{3}{5}+co{t}^{-1}\frac{3}{2}\right)$

(M) Let  $si{n}^{-1}\frac{3}{5}=x$ and cot^-1 $\frac{3}{2,}$ = y .

Then  $sinx=\frac{3}{5}coty=\frac{3}{2}$

Hence, tan x =  $\frac{sinx}{cosx}$ =  $\frac{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$5$}\right.}{\raisebox{1ex}{$4$}\!\left/ \!\raisebox{-1ex}{$5$}\right.}$ =  $\frac{3}{4}$

$tan\left(si{n}^{-1}\frac{3}{5}+co{t}^{-1}\frac{3}{2}\right)=tan(x+y)$

$=\frac{tanx+tany}{1-tanxtany}$ . $\frac{\frac{\mathrm{}}{}}{}$

$\frac{\frac{3\times 3+2\times 4}{4\times 3}}{\frac{4\times 3-3\times 2}{4\times 3}}$

$\frac{4\times 3-3\times 2}{4\times 3}$

$\frac{9+8}{12-6}=\frac{17}{6}$

Q19.  $\begin{array}{l}co{s}^{-1}\left(cos\frac{7\pi }{6}\right)isequalto\\ \left(A\right)\frac{7\pi }{6}\left(B\right)\frac{5\pi }{6}\left(C\right)\frac{\pi }{3}\left(D\right)\frac{\pi }{6}\end{array}$

A.19. Cos^-1 cos $\frac{7\pi }{6}$

(M) As  $\frac{7\pi }{6}$ [0, $\frac{\pi }{}$] ;principal value branch of cos^-1

cos^-1 $\left(cos\frac{7\pi }{6}\right)$ = cos^-1 $\left(cos2x-\frac{7\pi }{6}\right)$  $\{\because cos(2\pi -\theta )=cos\theta \}$

= cos^-1 $\left(sis\frac{12\pi -7\pi }{6}\right)$

$=co{s}^{-1}cos\frac{5\pi }{6}\frac{5\pi }{6}\in [0,\pi ]$

=  $\frac{5\pi }{6}$

So, option B is correct

Q20.  $\begin{array}{l}si{n}^{-1}\left(\frac{\pi }{3}-si{n}^{-1}\left(-\frac{1}{2}\right)\right)isequalto\\ \left(A\right)\frac{1}{2}\left(B\right)\frac{1}{3}\left(C\right)\frac{1}{4}\left(D\right)1\end{array}$

= sin  $\left[\frac{\pi }{3}+si{n}^{-1}\left(sin\frac{\pi }{6}\right)\right]$

= sin  $\left[\frac{\pi }{3}+\frac{\pi }{6}\right]$ = sin  $\left[\frac{2\pi +\pi }{6}\right]$ = sin  $\left(\frac{3\pi }{6}\right)$

= sin  $\frac{\pi }{2}=1$ .

The Miscellaneous Exercise of Inverse Trigonometric Function Chapter 2 Class 12 Maths contains 17 Questions (14 Short Answers, 3 MCQs). Students are provided with complete Miscellaneous Exercise solutions with detailed explanations including all the important concepts and more. The Chapter 2 ITF Miscellaneous Exercise consists of summarised examples of all the previous exercises. Students can complete Chapter 2 miscellaneous exercise well;

Inverse Trigonometric Function Miscellaneous Exercise Solutions

Find the value of the following:

Q1.  $co{s}^{-1}\left(cos\frac{13\pi }{6}\right)$

A.1. Cos^-1 $\left(cos\frac{13\pi }{6.}\right)$ =  $co{s}^{-1}$  $\left(cos\frac{12\pi +\pi }{6}\right)$

=  $co{s}^{-1}$  $\left(cos\frac{12\pi }{6}+\frac{\pi }{6}\right)$

=  $co{s}^{-1}$  $\left[co{s}^{-1}\left(2\pi +\frac{\pi }{6}\right)\right]$ {Øcor2 $\frac{\pi }{}$+=ØcosØ}

$=co{s}^{-1}\left(cos\frac{\pi }{6}\right)$

$=\frac{\pi }{6}$ ∈ [0,x]

Q2.  $ta{n}^{-1}\left(tan\frac{7\pi }{6}\right)$

A.2.  $ta{n}^{-1}\left(tan\frac{7\pi }{6}\right)=ta{n}^1\left(tan\frac{6\pi +\pi }{6}\right)$

$=ta{n}^{-1}\left(tan\frac{6\pi }{6}+\frac{\pi }{6}\right)$

$=ta{n}^{-1}\left(tan\pi +\frac{\pi }{6}\right)$

$=ta{n}^{-1}\left(tan\frac{\pi }{6}\right)$ { Ø tan ( $\frac{\pi }{}$+ Ø ) = tan Ø as tan b (+) we in 3rd quadrant)}

$=\frac{\pi }{6}$ ∈ $\left(\frac{-\pi }{2},\frac{\pi }{2}\right)$

Q4.  $2si{n}^{-1}\frac{8}{17}+si{n}^{-1}\frac{3}{5}=ta{n}^{-1}\frac{77}{36}$

A.4. Let  $si{n}^{-1}\frac{8}{17}=xsi{n}^{-1}\frac{3}{5}=y.$

(N. then,  $sinx=\frac{8}{17}siny=\frac{3}{5}.$

So,  $tanx=\frac{sinx}{cosx}$ and  $tany=\frac{siny}{cosy}$

$=\frac{8/17}{15/17}=\frac{3/4}{4/5}$

$=\frac{8}{15}=\frac{3}{4}$

Using.  $\text{\hspace{0.17em}tan}(x+y)=\frac{tanx+tany}{1-tanxtany}$

$tan\left(si{n}^{-1}\frac{8}{17}+si{n}^{-1}\frac{3}{5}\right)=\frac{\frac{8}{15}+\frac{3}{4}}{1-\frac{8}{15}\times \frac{3}{4}}$

tan  $(si{n}^{-1}\frac{8}{17}+si{n}^{-1}\frac{3}{5}=\frac{\frac{8\times 4+3\times 15}{15\times 4}}{\frac{15\times 4-8\times 3}{15\times 4}}=\frac{32+45}{60-24}\Rightarrow sin$

sin^-1 $\frac{8}{17}+si{n}^{-1}\frac{3}{5}=$ tan^-1 $\frac{77}{36}$

Hence proved.

Q5.  $co{s}^{-1}\frac{4}{5}+co{s}^{-1}\frac{12}{13}=co{s}^{-1}\frac{33}{65}$

A.5.  $co{s}^{-1}\frac{4}{5}+co{s}^{-1}\frac{12}{13}=co{s}^{-1}\frac{33}{65^{°}}.$

Let cos^-1 $\frac{4}{5}$ and cos^-1 $\frac{12}{13}$ = y.

Then,  $cosx=\frac{4}{5}\text{\hspace{0.17em}and}\text{\hspace{0.17em}cos}y=\frac{12}{13}.$

Using  $cos(x+y)=cosxcosy-sinxsiny.$

$\Rightarrow cos\left[co{s}^{-1}\frac{4}{5}+co{s}^{-1}\frac{12}{13}\right]=\frac{4}{5}\frac{12}{13}-\frac{3}{5}\times \frac{5}{13}$

$\Rightarrow cos\left[co{s}^{-1}\frac{4}{5}+co{t}^{-1}\frac{12}{13}\right]=\frac{48-15}{65}=\frac{33}{65}$

$\Rightarrow co{s}^{-1}\frac{4}{5}+co{s}^{-1}\frac{12}{13}=co{s}^{-1}\frac{33}{65}.\varphi$

Q6.  $co{s}^{-1}\frac{12}{13}+si{n}^{-1}\frac{3}{5}=si{n}^{-1}\frac{56}{65}$

A.6.  $co{s}^{-1}\frac{12}{13}+si{n}^{-1}\frac{3}{5}=si{n}^{-1}\frac{56}{65}$

Let  $co{s}^{-1}\frac{12}{13}=\mathrm{xandsin}{}^{-1}\frac{3}{5}=y.$

Thin,  $cosx=\frac{12}{13}\text{\hspace{0.17em}and sin}y=\frac{3}{5}$

Using  $sin(x+y)=sinxcosy+cosxsiny.$

$sin\left[co{s}^{-1}\frac{12}{13}+si{n}^{-1}\frac{3}{5}\right]=\frac{5}{13}\times \frac{4}{5}+\frac{12}{13}\times \frac{3}{5}=\frac{20+36}{65}=\frac{56}{65}.$

$co{s}^{-1}\frac{12}{13}+si{n}^{-1}\frac{3}{5}=si{n}^{-1}\frac{56}{65}.$

Q7.  $ta{n}^{-1}\frac{63}{16}=si{n}^{-1}\frac{5}{13}+co{s}^{-1}\frac{3}{5}$

A.7. Let  $si{n}^{-1}\frac{5}{13}=x\text{\hspace{0.17em}and}co{s}^{-1}\frac{3}{5}=y.$