# NCERT Solutions for Class 12 Maths Chapter 2 Inverse Trigonometric Functions - Exercise 2.2

### Access Exercises of Class 12 Maths Chapter 2 – Inverse Trigonometric Functions

Exercise 2.1 Solutions: 14 Questions (12 Short Answers, 2 MCQs)

Exercise 2.2 Solutions: 21 Questions (18 Short Answers, 3 MCQs)

Miscellaneous Exercise Solutions: 17 Questions (14 Short Answers, 3 MCQs)

Exercise 2.2

Prove the following :

$$\textbf{1.\space} \textbf{3}\space\textbf{sin}^{\textbf{-1}}\textbf{x}\space\textbf{=}\space\textbf{sin}^{\normalsize-1}\textbf{(3x-4x}^{\textbf{3}})\textbf{,}\\\textbf{x}\epsilon\bigg[-\frac{\textbf{1}}{\textbf{2}},\frac{\textbf{1}}{\textbf{2}}\bigg]\\\textbf{Sol.\space}\text{Let sin}^{-1}x=\theta\\\Rarr\space x=\text{sin}\space\theta,\text{then}$$

LHS = 3 sin–1x = 3 sin–1(sin θ) = 3θ

RHS = sin–1(3x – 4x3) = sin–1[3 sin θ – 4 sin3 θ]

= sin–1 [sin 3θ] = 3θ

[∵ sin 3θ = 3 sin θ – 4 sin3 θ]

∴ LHS = RHS Hence proved.

$$\textbf{2. 3 cos}^{\normalsize-1} \textbf{x = cos}^{\normalsize-1}\textbf{(4x}^{\textbf{3}}\textbf{- 3x})\textbf{,}\\ \textbf{x}\epsilon\bigg[\frac{\textbf{1}}{\textbf{2}}\textbf{,1}\bigg]$$

Sol. Let x = cos θ

LHS = 3 cos–1x = 3 cos–1<//sup>(cos θ) = 3θ

RHS = cos–1(4x3 – 3x) = cos–1[4 cos3 θ – 3 cos θ]

= cos–1 [cos 3θ] = 3θ

[∵ cos 3θ = 4 cos3 θ – 3 cos θ]

∴ LHS = RHS. Hence proved.

$$\textbf{3.\space}\textbf{tan}^{\normalsize-1}\frac{\textbf{2}}{\textbf{11}} + \textbf{tan}^{\normalsize-1}\frac{\textbf{7}}{\textbf{24}}\\\textbf{= tan}^{\normalsize-1}\frac{\textbf{1}}{\textbf{2}}\textbf{.}\\\textbf{Sol.\space}\text{LHS = tan}\\^{\normalsize-1}\frac{2}{11} + \text{tan}^{\normalsize-1}\frac{7}{24}\\=\text{tan}^{\normalsize-1}\bigg(\frac{\frac{2}{11} + \frac{7}{24}}{1-\frac{2}{11}×\frac{7}{24}}\bigg)\\=\text{tan}^{\normalsize-1}\bigg(\frac{\frac{48+7711}{11×24}}{\frac{11×24-14}{11×24}}\bigg)\\=\text{tan}^{\normalsize-1}\space\frac{48+77}{264-14}\\=\text{tan}^{\normalsize-1}\frac{125}{250}\\=\text{tan}^{\normalsize-1}\frac{1}{2}=\text{R.H.S}\space\\\textbf{Hence proved.}$$

$$\textbf{4.\space}\textbf{2 tan}^{\normalsize-1}\frac{\textbf{1}}{\textbf{2}} \textbf{+} \textbf{tan}^{\normalsize-1}\frac{\textbf{1}}{\textbf{7}}\textbf{=}\\\textbf{tan}^{\normalsize-1}\frac{\textbf{31}}{\textbf{17}}.\\\textbf{Sol.\space}\text{LHS = 2 tan}^{\normalsize-1}\frac{1}{2} + \text{tan}^{\normalsize-1}\frac{1}{7}\\= \text{tan}^{\normalsize-1}\begin{bmatrix}\frac{2×\frac{1}{2}}{1-\bigg(\frac{1}{2}\bigg)^{2}}\end{bmatrix}+\\\text{tan}^{\normalsize-1}\frac{1}{7}\\=\text{tan}^{\normalsize-1}\frac{1}{\bigg(\frac{3}{4}\bigg)}+\text{tan}^{\normalsize-1}\frac{1}{7}$$

$$=\text{tan}^{-1}\frac{4}{3} + \text{tan}^{\normalsize-1}\frac{1}{7}\\=\text{tan}^{\normalsize-1}\begin{pmatrix}\frac{\frac{4}{3}+\frac{1}{7}}{1-\frac{4}{3}×\frac{1}{7}}\end{pmatrix}\\=\text{tan}^{\normalsize-1}\frac{28+3}{21-4}\\=\text{tan}^{\normalsize-1}\frac{31}{16}=\text{RHS}$$

Write the following functions in the simplest form :

$$\textbf{5.\space tan}^{\normalsize-1}\frac{\sqrt{\textbf{1+x}^{\textbf{2}}}\textbf{-1}}{\textbf{x}}, \textbf{x}\neq\textbf{0}$$

Sol. Let x = tan θ, then θ = tan–1x

$$\therefore\space\text{tan}^{\normalsize-1}\frac{\sqrt{1 +x^{2}}-1}{x}\\=\text{tan}^{\normalsize-1}\frac{\sqrt{1 + \text{tan}^{2}\theta-1}}{\text{tan}\space\theta}\\=\text{tan}^{-1}\frac{\sqrt{\text{sec}^{2}\theta-1}}{\text{tan}\space\theta}\\\lbrack\because 1 + \text{tan}^{2}\theta = \text{sec}^{2}\theta\rbrack\\=\text{tan}^{\normalsize-1}\bigg[\frac{\text{sec}\space\theta-1}{\text{tan}\space\theta}\bigg]\\=\text{tan}^{\normalsize-1}\bigg[\frac{\frac{1}{\text{cos}\theta}-1}{\frac{\text{sin}\theta}{\text{cos}\space\theta}}\bigg]\\=\text{tan}^{\normalsize-1}\bigg[\frac{\frac{\text{1- cos}\theta}{\text{cos}\theta}}{\frac{\text{sin}\theta}{\text{cos}\theta}}\bigg]$$

$$\text{tan}^{\normalsize-1}\bigg[\frac{\text{1 - cos}\theta}{\text{cos}\theta}×\frac{\text{cos}\space\theta}{\text{sin}\space\theta}\bigg]\\=\text{tan}^{\normalsize-1}\bigg[\frac{\text{1 - cos}\space\theta}{\text{sin}\space\theta}\bigg]\\=\text{tan}^{-1}\bigg[\frac{2\text{sin}^{2}\frac{\theta}{2}}{2\text{sin}\frac{\theta}{2}\text{cos}\frac{\theta}{2}}\bigg]\\\begin{bmatrix}\because\space \text{1 - cos}\space\theta = 2\text{sin}^{2}\frac{\theta}{2}\\\text{and}\space\text{sin}\space\theta = 2\text{sin}\frac{\theta}{2}\text{cos}\frac{\theta}{2}\end{bmatrix}\\=\text{tan}^{\normalsize-1}\begin{bmatrix}\frac{\text{2 sin}\frac{\theta}{2}}{\text{2 cos}\frac{\theta}{2}}\end{bmatrix}\\=\text{tan}^{\normalsize-1}\bigg[\text{tan}\frac{\theta}{2}\bigg]=\frac{\theta}{2}\\=\frac{\text{tan}^{\normalsize-1}x}{2}$$

[from eq. (i)]

$$\therefore\space\text{tan}^{\normalsize-1}\frac{\sqrt{1 + x^{2}}-1}{x}\\=\frac{1}{2}\text{tan}^{\normalsize-1}x$$

$$\textbf{6. tan}^{\normalsize-1}\frac{\textbf{1}}{\sqrt{\textbf{x}^{\textbf{2}}\textbf{-1}}},\textbf{|x|}\gt\textbf{1.}$$

Sol. Let x = sec θ, then θ = sec–1x ...(i)

$$\therefore\space\text{tan}^{\normalsize-1}\frac{1}{\sqrt{x^{2}-1}}\\=\text{tan}^{\normalsize-1}\bigg(\frac{1}{\sqrt{\text{sec}^{2}\theta-1}}\bigg)\\=\text{tan}^{\normalsize-1}\bigg(\frac{1}{\sqrt{\text{tan}^{2}\theta}}\bigg)\\(\because\space\text{sec}^{2}\theta - \text{tan}^{2}\theta=1)\\=\text{tan}^{\normalsize-1}\bigg(\frac{1}{\text{tan}\space\theta}\bigg)\\=\text{tan}^{\normalsize-1}(\text{cot}\space\theta)\\=\text{tan}^{\normalsize-1}\bigg[\text{tan}\bigg(\frac{\pi}{2}-\theta\bigg)\bigg]\\=\bigg[\because\space\text{tan}\bigg(\frac{\pi}{2}-\theta\bigg) =\text{cot}\space\theta\bigg]$$

$$=\frac{\pi}{2}-\theta=\frac{\pi}{2}-\text{sec}^{\normalsize-1}x\space\\\lbrack\text{from Eq. (i)}\rbrack$$

$$\textbf{7.\space}\textbf{tan}^{\normalsize-1}\bigg(\sqrt{\frac{\textbf{1- cos x}}{\textbf{1+ cos x}}}\bigg)\textbf{,}\\\textbf{0}\lt \textbf{x}\lt\pi\textbf{.}\\\textbf{Sol.}\space\text{tan}^{\normalsize-1}\bigg(\frac{\text{1- cos x}}{\text{1 + cos x}}\bigg)\\=\text{tan}^{\normalsize-1}\bigg(\sqrt{\frac{2\text{sin}^{2}(x/2)}{\text{2 cos}^{2}(x/2)}}\bigg)\\\begin{pmatrix}\because\space \text{1 - cos x = 2 sin}^{2}(x/2)\\\text{and}\space\text{1 + cos x = 2 cos}^{2}(x/2)\end{pmatrix}\\=\text{tan}^{\normalsize-1}\bigg(\sqrt{\text{tan}^{2}\frac{x}{2}}\bigg)=\frac{x}{2}\\\textbf{Note\space:}\text{If x is positive then}\\\sqrt{\text{tan}^{2}\frac{x}{2}} = \text{tan}\frac{x}{2}$$

and if x is negative, then

$$\sqrt{\text{tan}^{2}\frac{x}{2}}=-\text{tan}\frac{x}{2}.$$

$$\textbf{8.\space}\textbf{tan}^{\normalsize-1}\bigg(\frac{\textbf{cos x - sin x}}{\textbf{cos x + sin x}}\bigg)\textbf{,}\\\frac{\pi}{\textbf{4}}\lt \textbf{x}\lt\frac{\textbf{3}\pi}{\textbf{4}}\\\textbf{Sol}\space\text{tan}^{\normalsize-1}\bigg(\frac{\text{cos x - sin x}}{\text{cos x + sin x}}\bigg)\\=\text{tan}^{\normalsize-1}\begin{pmatrix}\frac{\frac{\text{cos x}}{\text{cos x}} - \frac{\text{sin x}}{\textbf{cos x}}}{\frac{\text{cos x}}{cos x} + \frac{\text{sin x}}{\text{co sx}}}\end{pmatrix}$$

(inside the bracket divide numerator and denominator by cos x)

$$= \text{tan}^{\normalsize-1}\bigg(\frac{\text{1 - tan x}}{\text{1 + tan x}}\bigg)\\=\text{tan}^{\normalsize -1}\bigg[\text{{tan}}\bigg(\frac{\pi}{4}-x\bigg)\bigg]\\\begin{bmatrix}\because\tan\bigg(\frac{\pi}{4}-x\bigg) =\frac{\text{tan}\frac{\pi}{4} -\text{tan x}}{1 + \text{tan}\frac{\pi}{4}.\text{tan x}}\\\frac{\text{1 - tan x}}{\text{1 + tan x}}\end{bmatrix}\\=\frac{\pi}{4}-x$$

$$\textbf{9. tan}^{\normalsize-1}\frac{\textbf{x}}{\sqrt{\textbf{a}^{\textbf{2}}\textbf{-x}^{\textbf{2}}}}\textbf{,}|\textbf{x}|\lt \textbf{a.}$$

Sol. Let x = a sinθ

$$\Rarr\space\frac{x}{a} = \text{sin}\space\theta\\\Rarr\space\text{sin}^{\normalsize-1}\bigg(\frac{x}{a}\bigg)=\theta\space...\text{(i)}\\\therefore\space\text{tan}^{\normalsize-1}\frac{x}{\sqrt{a^{2}-x^{2}}}\\=\text{tan}^{\normalsize-1}\bigg(\frac{\text{a sin}\theta}{\sqrt{a^{2}-a^{2}\text{sin}^{2}\theta}}\bigg)\\=\text{tan}^{\normalsize-1}\bigg(\frac{\text{a sin}\theta}{a\sqrt{1 - \text{sin}^{2}\theta}}\bigg)\\=\text{tan}^{\normalsize-1}\bigg(\frac{\text{sin}\space\theta}{\text{cos}\space\theta}\bigg)\\\begin{pmatrix}\because\space\text{sin}^{2}x + \text{cos}^{2}x=1\\\Rarr\text{cos x} = \sqrt{1 - \text{sin}^{2}x}\end{pmatrix}$$

= tan−1 (tan θ) = θ

$$= \text{sin}^{\normalsize-1}\bigg(\frac{x}{a}\bigg)\\\lbrack\text{from Eq.(i)}\rbrack$$

$$\textbf{10. tan}^{\normalsize-1}\bigg(\frac{\textbf{3a}^{\textbf{2}}\textbf{x- x}^{\textbf{3}}}{\textbf{a}^{\textbf{3}} \textbf{- 3ax}^{\textbf{2}}}\bigg)\textbf{,}\\ \textbf{a}\gt\textbf{0;}\frac{\normalsize-\textbf{a}}{\sqrt{\textbf{3}}}\leq \textbf{x}\leq\frac{\textbf{a}}{\sqrt{\textbf{3}}}\textbf{.}$$

Sol. Let x = a tan θ

$$\Rarr\space\frac{x}{a} = \text{tan}\space\theta\\\Rarr\space\theta = \text{tan}^{\normalsize-1}\bigg(\frac{x}{a}\bigg)\\\text{...(1)}\\\therefore\space\text{tan}^{\normalsize-1}\bigg(\frac{3a^{2}x - x^{3}}{a^{3} - 3ax^{2}}\bigg)\\=\text{tan}^{\normalsize-1}\bigg(\frac{3a^{2}(a\text{tan}\space\theta) - (a\space\text{tan}\space\theta)^{3}}{a^{3}-3a(a\text{tan}\space\theta)^{2}}\bigg)\\=\text{tan}^{\normalsize-1}\bigg(\frac{a^{3}(3 \text{tan}\space\theta -\text{tan}^{3}\theta)}{a^{3}(1 - 3 \text{tan}^{2}\theta)}\bigg)\\=\text{tan}^{\normalsize-1}\bigg(\frac{\text{3 tan}\theta - \text{tan}^{3}\theta}{\text{1 - 3 tan}^{2}\theta}\bigg)$$

= tan–1 (tan 3θ)

$$\bigg(\because\space\text{tan}\space3\theta =\frac{\text{3 tan}\space\theta -\text{tan}^{3}\theta}{\text{1 - 3 tan}^{2}\theta}\bigg)\\= 3\theta = 3\text{tan}^{\normalsize-1}\bigg(\frac{x}{a}\bigg)\\\lbrack\text{from Eq.(i)}\rbrack$$

Find the value of each of the following questions :

$$\textbf{11.}\space\textbf{tan}^{\normalsize-1}\bigg[\textbf{2 cos}\bigg(\textbf{2 sin}^{\normalsize-1}\frac{\textbf{1}}{\textbf{2}}\bigg)\bigg]\textbf{.}\\\textbf{Sol.\space}\\\text{tan}^{\normalsize-1}\bigg[\text{2 cos}\bigg(\text{2 sin}^{\normalsize-1}\frac{1}{2}\bigg)\bigg]\\=\text{tan}^{\normalsize-1}\bigg[\text{2 cos}\begin{Bmatrix}\text{2 sin}^{\normalsize-1}\bigg(\text{sin}\frac{\pi}{6}\bigg)\end{Bmatrix}\bigg]\\\bigg(\because \text{sin}\frac{\pi}{6}=\frac{1}{2}\bigg)\\=\text{tan}^{\normalsize-1}\bigg[\text{2 cos}\bigg(2×\frac{\pi}{6}\bigg)\bigg]\\=\text{tan}^{\normalsize-1}\bigg[\text{2 cos}\frac{\pi}{3}\bigg]$$

$$= \text{tan}^{\normalsize-1}\bigg[2×\frac{1}{2}\bigg]\\=\text{tan}^{\normalsize-1}(1)\\\bigg(\because\space\text{cos}\frac{\pi}{3} =\frac{1}{2}\bigg)\\=\text{tan}^{\normalsize-1}\bigg(\text{tan}\frac{\pi}{4}\bigg)=\frac{\pi}{4}\\\bigg(\because\space\text{tan}\frac{\pi}{4}=1\bigg)$$

12. cot (tan–1 a + cot–1 a).

Sol. cot (tan–1 a + cot–1 a)

$$= \text{cot}\frac{\pi}{2}= 0\\\bigg(\because\space\text{tan}^{\normalsize-1}x + \text{cot}^{\normalsize-1} x=\frac{\pi}{2}\bigg)$$

$$\textbf{13.}\space\textbf{tan}\frac{\textbf{1}}{\textbf{2}}\begin{bmatrix}\textbf{sin}^{\normalsize-1}\frac{\textbf{2x}}{\textbf{1 + x}^{\textbf{2}}} + \\\textbf{cos}^{\normalsize-1}\frac{\textbf{1 - y}^{2}}{\textbf{1 + y}^{2}}\end{bmatrix}\textbf{,}\\\textbf{|x|}\lt \textbf{1, y}\gt \textbf{0}\space\textbf{and xy}\lt 1\textbf{.}\\\textbf{Sol.}\space\text{tan}\frac{1}{2}\bigg[\text{sin}^{\normalsize-1}\frac{2x}{1 +x^{2}} + \\\text{cos}^{\normalsize-1}\frac{\text{1 - y}^{2}}{\text{1 + y}^{2}}\bigg]\\\begin{bmatrix}\because\space\text{2 tan}^{\normalsize-1}x = \text{sin}^{\normalsize-1}\bigg(\frac{2x}{1 + x^{2}}\bigg)\\\text{and}\space 2\space\text{tan}^{\normalsize-1}y =\text{cos}^{\normalsize-1}\bigg(\frac{\text{1 - y}^{2}}{\textbf{1 + y}^{2}}\bigg)\end{bmatrix}\\=\text{tan}\bigg[\frac{1}{2}(2\space\text{tan}^{\normalsize-1}x + 2\space\text{tan}^{\normalsize-1}y)\bigg]$$

$$= \text{tan}\bigg[\frac{1}{2}× 2(\text{tan}^{\normalsize-1} x + \text{tan}^{\normalsize-1}y)\bigg]\\=\text{tan}(\text{tan}^{\normalsize-1}x + \text{tan}^{\normalsize-1}y)\\=\text{tan}\bigg[\text{tan}^{\normalsize-1}\bigg(\frac{x+y}{1-xy}\bigg)\bigg]\\=\frac{\text{x+y}}{\text{1 - xy}}\\\begin{bmatrix}\because\text{tan}^{\normalsize-1}x + \text{tan}^{\normalsize-1}y\\=\text{tan}^{\normalsize-1}\frac{x+y}{1-xy}\end{bmatrix}$$

$$\textbf{14.\space If sin}\bigg(\textbf{sin}^{\normalsize-1}\frac{\textbf{1}}{\textbf{5}} + \textbf{cos}^{\normalsize-1}\textbf{x =1,}\bigg)$$

then find the value of x.

$$\textbf{Sol.}\space\text{Given,}\\\text{sin}\bigg(\text{sin}^{\normalsize-1}\frac{1}{5} + \text{cos}^{\normalsize-1}x\bigg)=1\\\Rarr\space\text{sin}^{\normalsize-1}\frac{1}{5} + \text{cos}^{\normalsize-1}x =\text{sin}^{\normalsize-1} 1\\\lbrack\because\space\text{sin}\space\theta = x\\\Rarr\space\theta = \text{sin}^{\normalsize-1}x\rbrack\\\Rarr\space\text{sin}^{-1}\frac{1}{5} + \text{cos}^{\normalsize-1}x\\=\text{sin}^{\normalsize-1}\bigg(\text{sin}\frac{\pi}{2}\bigg)\\\bigg[\because\text{sin}\bigg(\frac{\pi}{2}\bigg)=1\bigg]$$

$$\Rarr\space\text{sin}^{\normalsize-1}\frac{1}{5} + \text{cos}^{\normalsize-1}x =\frac{\pi}{2}\\\Rarr\space \text{sin}^{-1}\frac{1}{5} =\frac{\pi}{2} -\text{cos}^{\normalsize-1}x\\\Rarr\space\text{sin}^{-1}\frac{1}{5} = \text{sin}^{\normalsize-1}x\\\bigg[\because\space\text{sin}^{\normalsize-1} x + \text{cos}^{\normalsize-1}x = \frac{\pi}{2}\bigg]\\\Rarr\space\frac{1}{5}=x$$

$$\textbf{15. If tan}^{\normalsize-1}\space\frac{\textbf{x-1}}{\textbf{x-2}} \textbf{+}\textbf{tan}^{\normalsize-1}\frac{\textbf{x+1}}{\textbf{x+2}}\textbf{=}\frac{\pi}{\textbf{4}}\textbf{,}$$

then find the value of x.

Sol. Given

$$\text{tan}^{\normalsize-1}\frac{x-1}{x-2} + \text{tan}^{\normalsize-1}\frac{x+1}{x+2}\\ =\frac{\pi}{4}\\\Rarr\space\text{tan}^{\normalsize-1}\begin{pmatrix}\frac{\frac{x-1}{x-2} + \frac{x+1}{x+2}}{1 - \bigg(\frac{x-1}{x-2}\bigg)\bigg(\frac{x+1}{x+2}\bigg)}\end{pmatrix}\\=\frac{\pi}{4}\\\begin{bmatrix}\because\space\text{tan}^{\normalsize-1}x + \text{tan}^{\normalsize-1}y\\=\text{tan}^{\normalsize-1}\bigg(\frac{x+y}{1-xy}\bigg)\end{bmatrix}\\\Rarr\text{tan}^{\normalsize-1}\begin{pmatrix}\frac{\frac{(x-1)(x+2) + (x+1)(x-2)}{2}}{\frac{(x-2)(x+2)-(x-1)(x+1)}{(x-2)(x+2)}}\end{pmatrix}\\=\frac{\pi}{4}$$

$$\Rarr\space\frac{(x-1)(x+2) + (x+1)(x-2)}{(x-2)(x+2) -(x-1)(x+1)}\\=\text{tan}\frac{\pi}{4}\\\Rarr\space\frac{2x^{2}-4}{(x^{2}-4) -(x^{2}-1)}=1\\\bigg(\because\text{tan}\frac{\pi}{4}=1\bigg)\\\Rarr\space 2x^{2}-4=-3\\\Rarr\space 2x^{2}=1\\\Rarr\space x^{2}=\frac{1}{2}\\\Rarr\space x=\pm\frac{1}{\sqrt{2}}$$

Find the values of each of the expressions.

$$\textbf{16.\space}\textbf{sin}^{\normalsize-1}\bigg(\textbf{sin}\frac{\textbf{2}\pi}{\textbf{3}}\bigg)\textbf{}.\\\textbf{Sol.\space}\text{sin}^{\normalsize-1}\bigg(\text{sin}\frac{2\pi}{3}\bigg)\\=\text{sin}^{\normalsize-1}\bigg[\text{sin}\bigg(\pi -\frac{\pi}{3}\bigg)\bigg]\\=\text{sin}^{\normalsize-1}\bigg(\text{sin}\frac{\pi}{3}\bigg)\\\lbrack\because\space \text{sin} (\pi-\theta) =\text{sin}\space\theta\rbrack\\=\frac{\pi}{3}\space\text{which lies in the range}\\\bigg[-\frac{\pi}{2},\frac{\pi}{2}\bigg].\\\textbf{Note : }\text{sin}^{\normalsize-1}(\text{sin}\space\theta) =\theta$$

$$\text{is only when θ}\space\epsilon\bigg[-\frac{\pi}{2},\frac{\pi}{2}\bigg].$$

$$\textbf{17.\space tan}^{\normalsize-1}\bigg(\textbf{tan}\frac{\textbf{3}\pi}{\textbf{4}}\bigg)\textbf{.}\\\textbf{Sol.\space}\text{tan}^{\normalsize-1}\bigg(\text{tan}\frac{3\pi}{4}\bigg)\\=\text{tan}^{\normalsize-1}\bigg[\text{tan}\bigg(\pi -\frac{\pi}{4}\bigg)\bigg]\\=\text{tan}^{\normalsize-1}\bigg(-\text{tan}\frac{\pi}{4}\bigg)\\\lbrack\because\space\text{tan}(\pi -\theta) = -\text{tan}\space\theta\rbrack\\=\text{tan}^{\normalsize-1}\bigg[\text{tan}\bigg(-\frac{\pi}{4}\bigg)\bigg]\\\lbrack\because\space -\text{tan}\space\theta =\text{tan}(-\theta)\rbrack\\=-\frac{\pi}{4}\space\text{which lies in the range}$$

$$\bigg(-\frac{\pi}{2},\frac{\pi}{2}\bigg).$$

$$\textbf{18. tan}\bigg(\textbf{sin}^{\normalsize-1}\frac{\textbf{3}}{\textbf{5}} + \textbf{cot}^{\normalsize-1}\frac{\textbf{3}}{\textbf{2}}\bigg)\\\textbf{Sol.\space}\text{tan}\bigg(\text{sin}^{\normalsize-1}\frac{3}{5} + \text{cot}^{\normalsize-1}\frac{3}{2}\bigg)\\\Rarr\space\text{tan}\begin{bmatrix}\text{tan}^{\normalsize -1}\frac{\frac{3}{5}}{\sqrt{1 -\bigg(\frac{3}{5}\bigg)^{2}}} + \text{tan}^{\normalsize-1}\frac{2}{3}\end{bmatrix}\\\begin{bmatrix}\because\text{sin}^{\normalsize-1}x =\text{tan}^{\normalsize-1}\frac{x}{\sqrt{1-x^{2}}}\\\text{and cot}^{\normalsize-1}\frac{x}{y} =\text{tan}^{\normalsize-1}\frac{y}{x}\end{bmatrix}\\=\text{tan}\bigg[\text{tan}^{\normalsize-1}\frac{3}{4} + \text{tan}^{\normalsize-1}\frac{2}{3}\bigg]$$

$$=\text{tan}\begin{bmatrix}\text{tan}^{\normalsize-1}\frac{\bigg(\frac{3}{4} + \frac{2}{3}\bigg)}{1 - \frac{3}{4}×\frac{2}{3}}\end{bmatrix}\\=\text{tan}\begin{bmatrix}\text{tan}^{\normalsize-1}\frac{\frac{17}{12}}{\frac{1}{2}}\end{bmatrix}\\\begin{bmatrix}\because\text{tan}^{\normalsize-1}x + \text{tan}^{\normalsize-1}y\\=\text{tan}^{\normalsize-1}\bigg(\frac{x+y}{1-xy}\bigg)\end{bmatrix}\\=\text{tan}\bigg[\text{tan}^{\normalsize-1}\frac{17}{6}\bigg]=\frac{17}{6}$$

$$\textbf{19.\space cos}^{\normalsize-1}\bigg(\textbf{cos}\frac{\textbf{7}\pi}{\textbf{6}}\bigg)\\\textbf{is equal to :}\\\textbf{(a)\space}\frac{\textbf{7}\pi}{\textbf{6}}\\\textbf{(b)\space}\frac{\textbf{5}\pi}{\textbf{6}}\\\textbf{(c)\space}\frac{\pi}{\textbf{3}}\\\textbf{(d)\space}\frac{\pi}{6}\\\textbf{Sol.\space}\text{(b)}\frac{5\pi}{6}$$

$$\text{cos}^{\normalsize-1}\bigg(\text{cos}\frac{7\pi}{6}\bigg)\\=\text{cos}^{\normalsize-1}\bigg[\text{cos}\bigg(2\pi -\frac{5\pi}{6}\bigg)\bigg]\\\text{where,}\space\frac{5\pi}{6}\epsilon\lbrack0,\pi\rbrack\\\therefore\space\text{cos}^{\normalsize-1}\bigg(\text{cos}\frac{7\pi}{6}\bigg) \\=\text{cos}^{\normalsize-1}\bigg[\text{cos}\bigg(\frac{5\pi}{6}\bigg)\bigg]=\frac{5\pi}{6}\\\lbrack\because\space \text{cos}(2\pi -\theta) = \text{cos}\space\theta\rbrack$$

$$\textbf{20.\space}\textbf{sin}\bigg[\frac{\pi}{\textbf{3}} \textbf{-}\textbf{sin}^{\normalsize-1}\bigg(\textbf{-}\frac{\textbf{1}}{\textbf{2}}\bigg)\bigg]\\\textbf{is equal to}\\\textbf{(a)}\space\frac{\textbf{1}}{\textbf{2}}\\\textbf{(b)\space}\frac{\textbf{1}}{\textbf{3}}\\\textbf{(c)\space}\frac{\textbf{1}}{\textbf{4}}$$

(d) 1

Sol. (d) 1

$$\text{sin}\bigg[\frac{\pi}{3} -\text{sin}^{\normalsize-1}\bigg(-\frac{1}{2}\bigg)\bigg]\\=\text{sin}\bigg[\frac{\pi}{3} -\text{sin}^{\normalsize-1}\bigg(-\text{sin}\frac{\pi}{6}\bigg)\bigg]\\\bigg[\because\text{sin}\frac{\pi}{6} =\frac{1}{2}\bigg]\\=\text{sin}\bigg[\frac{\pi}{3} -\text{sin}^{\normalsize-1}\begin{Bmatrix}\text{sin}\bigg(-\frac{\pi}{6}\bigg)\end{Bmatrix}\bigg]\\\lbrack\because\space \text{sin}(-x) =-\text{sin x}\rbrack\\=\text{sin}\bigg[\frac{\pi}{3} -\bigg(\frac{\pi}{6}\bigg)\bigg]\\=\text{sin}\bigg[\frac{\pi}{3}+\frac{\pi}{6}\bigg]\\=\text{sin}\frac{\pi}{2} = 1$$

$$\textbf{21. tan}^{\normalsize-1}\sqrt{\textbf{3}}\space\textbf{- cot}^{\normalsize-1}(-\sqrt{\textbf{3}})\\\textbf{is equal to}\\\textbf{(a)\space}\pi\\\textbf{(b)\space}-\frac{\pi}{\textbf{2}}\\\textbf{(c)\space} 0\\\textbf{(d)\space}\textbf{2}\sqrt{\textbf{3}}\\\textbf{Sol.}(b)\space-\frac{\pi}{2}\\\text{tan}^{\normalsize-1}\sqrt{3} -\text{cot}^{-1}(-\sqrt{3})\\=\text{tan}^{\normalsize-1}\sqrt{3} - \lbrack\pi -\text{cot}^{-1}\sqrt{3}\rbrack\\\lbrack\because\text{cot}^{\normalsize-1}(-x) =\pi -\text{cot}^{\normalsize-1}x\rbrack\\=\text{tan}^{\normalsize-1}\sqrt{3} - \pi +\text{cot}^{\normalsize-1}\sqrt{3}\\=\text{tan}^{\normalsize-1}\sqrt{3}-\pi +\text{tan}^{\normalsize-1}\frac{1}{\sqrt{3}}$$

$$\bigg(\because\text{tan}^{\normalsize-1}\frac{1}{x} =\text{cot}^{\normalsize-1}x\bigg)\\=\bigg[\text{tan}^{\normalsize-1}\sqrt{3} + \text{tan}^{\normalsize-1}\frac{1}{\sqrt{3}}\bigg]\\-\pi\\=\text{tan}^{\normalsize-1}\frac{\bigg(\sqrt{3} + \frac{1}{\sqrt{3}}\bigg)}{1-\bigg(\sqrt{3}×\frac{1}{\sqrt{3}}\bigg)}-\pi\\\begin{bmatrix}\because\space\text{tan}^{\normalsize-1}x + \text{tan}^{-1}y\\\text{tan}^{\normalsize-1}\bigg(\frac{x+y}{1-xy}\bigg)\end{bmatrix}\\=\text{tan}^{\normalsize-1}(\infty)-\pi$$

$$=\frac{\pi}{2}-\pi =-\frac{\pi}{2}\\\bigg(\because\text{tan}\frac{\pi}{2} =\infty\bigg)$$