Inverse Trigonometric Functions - Class 12 Commerce Maths - Extra Questions

We know that the cosine function, is nothing more than the sine $$\dfrac{\pi}{2} $$ radians out of phase, as proved below:

$$cos\left (\theta- \dfrac{\pi}{2} \right ) = cos\left (\theta \right )cos\left (- \dfrac{\pi}{2} \right ) - sin\left (\theta \right )sin\left (- \dfrac{\pi}{2} \right )$$

$$cos\left (\theta- \dfrac{\pi}{2} \right ) = cos\left (\theta \right )\times 0 - \left ( sin\left (\theta \right )sin\left (\dfrac{\pi}{2} \right ) \right )$$

$$cos\left (\theta- \dfrac{\pi}{2} \right ) = sin\left (\theta \right ) \times 1 = sin\left (\theta \right )$$

It's clear that

$$cos\left (\theta \right ) = sin \left (x + \dfrac{\pi}{2}  \right )$$

$$sin^{-1} \left ( sin \left (x + \dfrac{\pi}{2}  \right ) \right )$$

$$= \left (x + \dfrac{\pi}{2}  \right )$$

$${\left( {{{\tan }^{ - 1}}x} \right)^2} + \left( {{{\cot }^{ - 1}}x} \right) = {{5{\pi ^2}} \over 8}$$

                $${\left( {{{\tan }^{ - 1}}x} \right)^2} + \left( {{\pi  \over 2} - {{\tan }^{ - 1}}{x^2}} \right) = {{5{\pi ^2}} \over 8}$$

                Let $${\tan ^{ - 1}}x = k$$

                Then $${k^2} + {\left( {{\pi  \over 2} - k} \right)^2} = {{5{\pi ^2}} \over 8}$$

                $${k^2} + {{{\pi ^2}} \over 4} + {k^2} - \pi k = {{5{\pi ^2}} \over 8}$$

                $$ = 2{k^2} - \pi k + {{{\pi ^2}} \over 4} - {{5{\pi ^2}} \over 8} = 0$$

                $$ = 2{k^2} - \pi k - {{3{\pi ^2}} \over 8} = 0$$

                $$ = 16{k^2} - 8\pi k - 3{\pi ^2} = 0$$

                $$ = \left( {4k - 3\pi } \right)\left( {4k + \pi } \right)$$

                $$k = {{3\pi } \over 4},k = {{ - \pi } \over 4}$$

                $$x = \tan \left( {{{3\pi } \over 4}} \right);x = \tan \left( { - {\pi _4}} \right)$$

                $$x =  - 1$$

L.H.S

$$\cot \left( \dfrac{\pi }{2}-2{{\cot }^{-1}}3 \right)$$

We know that

$$\cot \left( \dfrac{\pi }{2}-\theta  \right)=\tan \theta $$

Therefore,

$$ \Rightarrow \tan \left( 2{{\cot }^{-1}}3 \right) $$

$$ \Rightarrow \tan \left( 2{{\tan }^{-1}}\left( \dfrac{1}{3} \right) \right)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left[ \because {{\cot }^{-1}}x={{\tan }^{-1}}\left( \dfrac{1}{x} \right) \right] $$

$$ \Rightarrow \tan \left( {{\tan }^{-1}}\left( \dfrac{2\times \dfrac{1}{3}}{1-{{\left( \dfrac{1}{3} \right)}^{2}}} \right) \right)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left[ 2{{\tan }^{-1}}x={{\tan }^{-1}}\left( \dfrac{2x}{1-{{x}^{2}}} \right) \right] $$

$$ \Rightarrow \dfrac{\dfrac{2}{3}}{1-\dfrac{1}{9}} $$

$$ \Rightarrow \dfrac{\dfrac{2}{3}}{\dfrac{8}{9}} $$

$$ \Rightarrow \dfrac{2\times 9}{3\times 8} $$

$$ \Rightarrow \dfrac{3}{4} $$

Hence, this is the answer.

$$sec^{-1}(-2)\\=cos^{-1}((\frac{-1}{2}))\\=\pi-cos^{-1}((\frac{1}{2}))\\=\pi -(\frac{\pi}{3})\\=(\frac{2\pi}{3})$$

Given that,$${\tan ^{ - 1}}\left( {\tan \dfrac{{2\pi }}{3}} \right)$$ ……….     (1)

First solve that,

$$\tan \left( \dfrac{2\pi }{3} \right)$$

$$ =\tan \left( \pi -\dfrac{\pi }{3} \right) $$

$$ =\tan \left( \dfrac{-\pi }{3} \right) $$

$$ =-\tan \dfrac{\pi }{3} $$

$$ =-\sqrt{3} $$

By equation (1),

$$=\tan \left( -\sqrt{3} \right)$$

Let

$$ \tan \left( -\sqrt{3} \right)=x $$

$$ \tan x=-\sqrt{3} $$

Then,$$-\dfrac{\pi }{2}\le x\le \dfrac{\pi }{2}$$

Hence, this is the answer.

Consider the given function  $$={{\sin }^{-1}}\left( \dfrac{1-{{x}^{2}}}{1+{{x}^{2}}} \right)$$

Let, put $$x=\tan A$$       then    $$A={{\tan }^{-1}}x$$ …….   (1)

Then,  

$$ {{\sin }^{-1}}\left( \dfrac{1-{{\tan }^{2}}A}{1+{{\tan }^{2}}A} \right) $$

$$ ={{\sin }^{-1}}\left( \dfrac{1-\dfrac{{{\sin }^{2}}A}{{{\cos }^{2}}A}}{1+\dfrac{{{\sin }^{2}}A}{{{\cos }^{2}}A}} \right) $$

$$ ={{\sin }^{-1}}\left( \dfrac{{{\cos }^{2}}A-{{\sin }^{2}}A}{{{\cos }^{2}}A+{{\sin }^{2}}A} \right) $$

$$ ={{\sin }^{-1}}\left( \dfrac{\cos 2A}{1} \right) $$

$$ ={{\sin }^{-1}}\cos 2A $$

$$ ={{\sin }^{-1}}\sin \left( {{90}^{0}}-2A \right) $$

$$ ={{90}^{0}}-2A $$

By equation (1)  and we get,

$$={{90}^{0}}-2{{\tan }^{-1}}x$$

Hence, this is the answer.

We have,

$$ {{\sin }^{-1}}\sin {{15}^{o}}+{{\cos }^{-1}}\cos {{20}^{o}}+{{\tan }^{-1}}\tan {{25}^{o}} $$

$$ \Rightarrow {{15}^{o}}+{{20}^{o}}+{{25}^{o}} $$

$$ \Rightarrow {{60}^{o}}=\dfrac{\pi }{3} $$

$$LHS=\tan^{-1}\dfrac{1}{2}+\tan^{-1}\dfrac{2}{11}\\=\tan^{-1}[(\dfrac{\dfrac{1}{2}+\dfrac{2}{11}}{1-(\dfrac{1}{2})(\dfrac{2}{11})})]\\=\tan^{-1}(\dfrac{(\dfrac{11+4}{22})}{(\dfrac{22-2}{22})})\\=\tan^{-1}\dfrac{15}{20}=\tan^{-1}\dfrac{3}{4}=RHS$$

Let$$\>x\>=\>tan\theta\>or\>\theta\>=\>tan^{-1}x\\  =\> sin^{-1}(\dfrac{2tan\theta\>}{1+tan^2\theta\>})\\=sin^{-1}(sin2\theta)\\=2\theta\\=2tan^{-1}x$$

Differentiating w.r.t x, We get:

$$\>(\dfrac{2}{1+x^2})$$

Consider the given expression,

$$ \Rightarrow {{\tan }^{-1}}\left( \tan 5 \right) $$

$$ \Rightarrow 5 $$

Hence, this is the answer.

We have,

$$ \sin {{\cot }^{-1}}\cot x $$

$$ \Rightarrow \sin x $$

Let $$\cos A=x $$ then $$\cos(\pi-A)=-\cos A=-x$$

Hence, $$\cos^{-1}x=A$$ 

$$\cos^{-1}(-x)=\pi-A$$ 

Adding them,

$$\cos^{-1}x+\cos^{-1}(-x)=A+\pi-A\Rightarrow \cos^{-1}x+\cos^{-1}(-x)=\pi\\ \Rightarrow \cos^{-1}(-x)=\pi-\cos^{-1}(x)$$

Given that

$$(tan^{-1}x)^2+(cot^{-1}x)^2=\dfrac{5\pi^2}{8}$$

We know that,

$$tan^{-1}x+cot^{-1}x=\dfrac{\pi}{2}$$                      .....$$(1)$$

Squaring on both sides

$$\Rightarrow (tan^{-1}x+cot^{-1}x)^2=(\dfrac{\pi}{2})^2$$

$$\Rightarrow (tan^{-1}x)^2+(cot^{-1}x)^2+2(tan^{-1}x)(cot^{-1}x)=\dfrac{\pi^2}{4}$$      ....$$(2)$$

from the given information,

$$(tan^{-1}x)^2+(cot^{-1}x)^2=\dfrac{5\pi^2}{8}$$

Substituting this in $$(2)$$, we get

$$\Rightarrow \dfrac{5\pi^2}{8}+2(tan^{-1}x)(cot^{-1}x)=\dfrac{\pi^2}{4}$$

$$\Rightarrow 2(tan^{-1}x)(cot^{-1}x)=\dfrac{-3\pi^2}{8}$$             ....$$(3)$$         

We know that $$(a-b)^2=(a+b)^2-4ab$$

Using this formula, we obtain

$$(tan^{-1}x-cot^{-1}x)^2=(tan^{-1}x+cot^{-1}x)^2-4(tan^{-1}x)(cot^{-1}x)$$

$$\Rightarrow (tan^{-1}x-cot^{-1}x)^2=\dfrac{\pi^2}{4}-2(\dfrac{-3\pi^2}{8})$$                  $$(from (2),(3))$$

$$\Rightarrow (tan^{-1}x-cot^{-1}x)^2=\dfrac{\pi^2}{4}+\dfrac{3\pi^2}{4}$$

$$(tan^{-1}x-cot^{-1}x)^2=\pi^2$$

Taking square root,

$$\Rightarrow tan^{-1}x-cot^{-1}x=\pi$$                  .....$$(4)$$

On solving $$(1)$$ and $$(4)$$, we get

$$\Rightarrow 2tan^{-1}x=\dfrac{3\pi}{2}$$

$$\Rightarrow tan^{-1}x=\dfrac{3\pi}{4}$$

$$\Rightarrow x=tan(\dfrac{3\pi}{4})$$

$$\Rightarrow x=-1 $$

Let,

$$\dfrac{17\pi^2}{36}=(sin^{-1}x+cos^{-1}x)^2-2sin^{-1}xcos^{-1}x$$     ($$a^2+b^2=(a+b)^2-2ab$$)

$$\dfrac{{17{\pi ^2}}}{{36}}=\dfrac{\pi^2}{4}-2sin^{-1}[\dfrac{\pi}{2}-sin^{-1}x]$$

$$(sin^{-1})^2-\dfrac{\pi}{2}(sin^{-1}x)-\dfrac{\pi^2}{9}=0$$

$$sin^{-1}x=\dfrac{\dfrac{\pi}{2} \pm \sqrt{\dfrac{\pi^2}{4}+\dfrac{4\pi^2}{9}}}{2 (1)}$$

$$sin^{-1}x=\dfrac{\pi}{2},-\dfrac{\pi}{6}$$

$$sin^{-1}x=-\dfrac{\pi}{6}$$

$$x=-\dfrac{1}{2}$$

$$let x=sin\theta \> then\> \theta =sin^{-1}x\\then \>sin^{-1}(2sin\theta \sqrt{1-sin^2\theta })\\=sin^{-1}(2sin\theta cos\theta )\\=sin^{-1}(sin2\theta )\\=2\theta \\=2sin^{-1}x$$

We have,

$$\tan \left\{ \dfrac{1}{2}{{\sin }^{-1}}\left( \dfrac{2x}{1+{{x}^{2}}} \right)+\dfrac{1}{2}{{\cos }^{-1}}\left( \dfrac{1-{{y}^{2}}}{1+{{y}^{2}}} \right) \right\}$$

Put $$x=y=\tan \theta $$

Then, $$\theta ={{\tan }^{-1}}x={{\tan }^{-1}}y$$

$$ \tan \left\{ \dfrac{1}{2}{{\sin }^{-1}}\left( \dfrac{2x}{1+{{x}^{2}}} \right)+\dfrac{1}{2}{{\cos }^{-1}}\left( \dfrac{1-{{y}^{2}}}{1+{{y}^{2}}} \right) \right\} $$

$$ =\tan \left\{ \dfrac{1}{2}{{\sin }^{-1}}\left( \dfrac{2\tan \theta }{1+{{\tan }^{2}}\theta } \right)+\dfrac{1}{2}{{\cos }^{-1}}\left( \dfrac{1-{{\tan }^{2}}\theta }{1+{{\tan }^{2}}\theta } \right) \right\} $$

$$ =\tan \left\{ \dfrac{1}{2}{{\sin }^{-1}}\sin 2\theta +\dfrac{1}{2}{{\cos }^{-1}}\cos 2\theta  \right\} $$

$$ =\tan \left\{ \dfrac{1}{2}\times 2\theta +\dfrac{1}{2}\times 2\theta  \right\} $$

$$ =\tan 2\theta  $$

$$ =\dfrac{2\tan \theta }{1-{{\tan }^{2}}\theta } $$

$$ =\dfrac{2x}{1-{{x}^{2}}}=\dfrac{2y}{1-{{y}^{2}}} $$

We have,

$${{\tan }^{-1}}\left( 1 \right)+{{\cos }^{-1}}\left( -\dfrac{1}{2} \right)+{{\sin }^{-1}}\left( -\dfrac{1}{2} \right)$$

$$ \Rightarrow {{\tan }^{-1}}\left( \tan \dfrac{\pi }{4} \right)+{{\cos }^{-1}}\left( \cos \dfrac{2\pi }{3} \right)+{{\sin }^{-1}}\left( \sin \dfrac{7\pi }{6} \right) $$

$$ \Rightarrow \dfrac{\pi }{4}+\dfrac{2\pi }{3}+\dfrac{7\pi }{6} $$

$$ \Rightarrow \dfrac{3\pi +8\pi +14\pi }{12} $$

$$ \Rightarrow \dfrac{25\pi }{12} $$

Hence, this is the answer.

$$tan^{-1}(\frac{(\frac{1}{3})+(\frac{1}{7})}{1-(\frac{1}{21})})+tan^{-1}(\frac{(\frac{1}{5})+(\frac{1}{8})}{1-(\frac{1}{40})})\\=tan^{-1}(\frac{10}{20})+tan^{-1}(\frac{13}{39})\\=tan^{-1}(\frac{1}{2})+tan^{-1}(\frac{1}{3})\\=tan^{-1}(\frac{(\frac{1}{2})+(\frac{1}{3})}{1-(\frac{1}{20})})\\tan^{-1}(\frac{5}{5})=tan^{-1}1=(\frac{\pi}{4})$$

$$cos^{-1}(\frac{-1}{2})=\pi-cos^{-1}(\frac{1}{2})\\=\pi-(\frac{\pi}{3})=(\frac{2\pi}{3})$$

We have,

$$ \dfrac{{{a}^{2}}-{{b}^{2}}}{4}{{\sin }^{-1}}\dfrac{a-b}{\sqrt{{{a}^{2}}-{{b}^{2}}}} $$

$$ \Rightarrow \dfrac{{{a}^{2}}-{{b}^{2}}}{4}{{\sin }^{-1}}\dfrac{a-b}{\sqrt{\left( a+b \right)\left( a-b \right)}} $$

$$ \Rightarrow \dfrac{{{a}^{2}}-{{b}^{2}}}{4}{{\sin }^{-1}}\dfrac{\sqrt{a-b}\sqrt{a-b}}{\sqrt{a+b}\sqrt{a-b}} $$

$$ \Rightarrow \dfrac{{{a}^{2}}-{{b}^{2}}}{4}{{\sin }^{-1}}\dfrac{\sqrt{a-b}}{\sqrt{a+b}} $$

Hence, this is the answer.

We have,

$${{\sin }^{-1}}\left( \dfrac{\sqrt{1+x}+\sqrt{1-x}}{2} \right)$$

Let put $$x=\cos 2\theta $$

So,

$$ {{\sin }^{-1}}\left( \dfrac{\sqrt{1+\cos 2\theta }+\sqrt{1-\cos 2\theta }}{2} \right) $$

$$ \Rightarrow {{\sin }^{-1}}\left( \dfrac{\sqrt{1+2{{\cos }^{2}}\theta -1}+\sqrt{1-1+2{{\sin }^{2}}\theta }}{2} \right) $$

$$ \Rightarrow {{\sin }^{-1}}\left( \dfrac{\sqrt{2{{\cos }^{2}}\theta }+\sqrt{2{{\sin }^{2}}\theta }}{2} \right) $$

$$ \Rightarrow {{\sin }^{-1}}\left( \dfrac{\cos \theta +\sin \theta }{\sqrt{2}} \right) $$

$$ \Rightarrow {{\sin }^{-1}}\left( \dfrac{\cos \theta }{\sqrt{2}}+\dfrac{\sin \theta }{\sqrt{2}} \right) $$

$$ \Rightarrow {{\sin }^{-1}}\left( sin{{45}^{o}}\cos \theta +cos{{45}^{o}}\sin \theta  \right) $$

$$ \Rightarrow {{\sin }^{-1}}\sin \left( {{45}^{o}}+\theta  \right) $$

$$ \Rightarrow {{45}^{o}}+\theta  $$

$$ \Rightarrow {{45}^{o}}+\dfrac{1}{2}{{\cos }^{-1}}x $$

Hence, this is the answer.

We have,

$$y={{\cot }^{-1}}\dfrac{2x}{1-{{x}^{2}}}$$

Let put $$x=\tan \theta $$

Then

$$\theta ={{\tan }^{-1}}x$$

$$ y={{\cot }^{-1}}\dfrac{2\tan \theta }{1-{{\tan }^{2}}\theta } $$

$$ y={{\cot }^{-1}}\tan 2\theta \,\,\,\,\,\,\,\left( \therefore \tan 2\theta =\dfrac{2\tan \theta }{1-{{\tan }^{2}}\theta } \right) $$

$$ y={{\cot }^{-1}}\cot \left( \dfrac{\pi }{2}-2\theta  \right) $$

$$ y=\dfrac{\pi }{2}-2\theta  $$

$$ y=\dfrac{\pi }{2}-2{{\tan }^{-1}}x $$

We have,

$${{\tan }^{-1}}\dfrac{1}{2}+{{\tan }^{-1}}\dfrac{1}{3}$$

We know that,

$${{\tan }^{-1}}x+{{\tan }^{-1}}y={{\tan }^{-1}}\left( \dfrac{x+y}{1-xy} \right)$$

So,

$$ {{\tan }^{-1}}\dfrac{1}{2}+{{\tan }^{-1}}\dfrac{1}{3}={{\tan }^{-1}}\left( \dfrac{\dfrac{1}{2}+\dfrac{1}{3}}{1-\dfrac{1}{2}\times \dfrac{1}{3}} \right) $$

$$ ={{\tan }^{-1}}\left( \dfrac{\dfrac{3+2}{6}}{1-\dfrac{1}{6}} \right) $$

$$ ={{\tan }^{-1}}\left( \dfrac{5}{5} \right) $$

$$ ={{\tan }^{-1}}1 $$

$$ ={{\tan }^{-1}}\tan \dfrac{\pi }{4} $$

$$ =\dfrac{\pi }{4} $$

We have,

$${{\tan }^{-1}}\left( \dfrac{a\cos x-b\sin x}{b\cos x+a\sin x} \right)$$

On divide numerator and denominator by cosx and we get,

$$ {{\tan }^{-1}}\left( \dfrac{a\cos x-b\sin x}{b\cos x+a\sin x} \right) $$

$$ \Rightarrow {{\tan }^{-1}}\left( \dfrac{\dfrac{a\cos x-b\sin x}{\cos x}}{\dfrac{b\cos x+a\sin x}{\cos x}} \right) $$

$$ \Rightarrow {{\tan }^{-1}}\left( \dfrac{\dfrac{a\cos x}{\cos x}-\dfrac{b\sin x}{\cos x}}{\dfrac{b\cos x}{\cos x}+\dfrac{a\sin x}{\cos x}} \right) $$

$$ \Rightarrow {{\tan }^{-1}}\left( \dfrac{a-b\operatorname{tanx}}{b+a\tan x} \right) $$

$$ \Rightarrow {{\tan }^{-1}}\dfrac{b}{b}\left( \dfrac{\dfrac{a}{b}-\tan x}{1+\dfrac{a}{b}\tan x} \right) $$

$$ \Rightarrow {{\tan }^{-1}}\left( \dfrac{\tan y-\tan x}{1+\tan y\tan x} \right)\,\,\,\,\,\therefore let\,\,\tan y\,=\dfrac{a}{b} $$

$$ \Rightarrow {{\tan }^{-1}}\tan \left( y-x \right)\,\,\,\,\,\therefore So,\,y\,={{\tan }^{-1}}\dfrac{a}{b} $$

$$ \Rightarrow \left( y-x \right) $$

$$ \Rightarrow {{\tan }^{-1}}\dfrac{a}{b}-x $$

Hence, this is the answer.

We have,

$${{\cos }^{-1}}x+{{\cos }^{-1}}y+{{\cos }^{-1}}z=\pi \,\,\,.....\,\,\left( 1 \right)$$

$$ {{\cos }^{-1}}x+{{\cos }^{-1}}y=\pi -{{\cos }^{-1}}z $$

$$ \Rightarrow {{\cos }^{-1}}\left[ xy-\sqrt{\left( 1-{{x}^{2}} \right)\left( 1-{{y}^{2}} \right)} \right]=\pi -{{\cos }^{-1}}z $$

$$ \therefore \,\,{{\cos }^{-1}}A+{{\cos }^{-1}}B={{\cos }^{-1}}\left[ AB-\sqrt{\left( 1-{{A}^{2}} \right)\left( 1-{{B}^{2}} \right)} \right] $$

$$ \Rightarrow \left[ xy-\sqrt{\left( 1-{{x}^{2}} \right)\left( 1-{{y}^{2}} \right)} \right]=\cos \left( \pi -{{\cos }^{-1}}z \right) $$

$$ \Rightarrow xy-\sqrt{\left( 1-{{x}^{2}} \right)\left( 1-{{y}^{2}} \right)}=-\cos {{\cos }^{-1}}z $$

$$ \Rightarrow xy-\sqrt{\left( 1-{{x}^{2}} \right)\left( 1-{{y}^{2}} \right)}=-z $$

$$ \Rightarrow xy+z=\sqrt{\left( 1-{{x}^{2}} \right)\left( 1-{{y}^{2}} \right)} $$

On squaring both side and we get,

$$ {{\left( xy+z \right)}^{2}}=\left( 1-{{x}^{2}} \right)\left( 1-{{y}^{2}} \right) $$

$$ \Rightarrow {{x}^{2}}{{y}^{2}}+{{z}^{2}}+2xyz=1-{{x}^{2}}-{{y}^{2}}+{{x}^{2}}{{y}^{2}} $$

$$ \Rightarrow {{x}^{2}}+{{y}^{2}}+{{z}^{2}}+2xyz=1 $$

Hence proved.

We have,

$$ {{\left( {{\tan }^{-1}}x \right)}^{2}}+{{\left( {{\tan }^{-1}}x \right)}^{2}}=\dfrac{5{{\pi }^{2}}}{8} $$

$$ \Rightarrow 2{{\left( {{\tan }^{-1}}x \right)}^{2}}=\dfrac{5{{\pi }^{2}}}{8} $$

$$ \Rightarrow {{\left( {{\tan }^{-1}}x \right)}^{2}}=\dfrac{5{{\pi }^{2}}}{16} $$

$$ \Rightarrow {{\tan }^{-1}}x=\sqrt{\dfrac{5{{\pi }^{2}}}{16}} $$

$$ \Rightarrow {{\tan }^{-1}}x=\sqrt{5}\dfrac{\pi }{4} $$

$$ \Rightarrow x=\tan \dfrac{\sqrt 5 \pi }{4} $$

We have,

$${{\sin }^{-1}}x=\dfrac{\pi }{5}$$

We know that,

$${{\sin }^{-1}}x+{{\cos }^{-1}}x=\dfrac{\pi }{2}$$

Now,

$$ {{\sin }^{-1}}x+{{\cos }^{-1}}x=\dfrac{\pi }{2} $$

$$ \Rightarrow \dfrac{\pi }{5}+{{\cos }^{-1}}x=\dfrac{\pi }{2} $$

$$ \Rightarrow {{\cos }^{-1}}x=\dfrac{\pi }{2}-\dfrac{\pi }{5} $$

$$ \Rightarrow {{\cos }^{-1}}x=\dfrac{5\pi -2\pi }{10} $$

$$ \Rightarrow {{\cos }^{-1}}x=\dfrac{3\pi }{10} $$

Hence, this is the answer.

We have,

$$ {{\cos }^{-1}}x+{{\cos }^{-1}}y+{{\cos }^{-1}}z=\pi \,\,......\,\,\left( 1 \right) $$

$$ \Rightarrow {{\cos }^{-1}}x+{{\cos }^{-1}}y=\pi -{{\cos }^{-1}}z $$

We know that,

$${{\cos }^{-1}}A+{{\cos }^{-1}}B={{\cos }^{-1}}\left[ AB-\sqrt{1-{{A}^{2}}}\sqrt{1-{{B}^{2}}} \right]$$

Therefore,

$$ {{\cos }^{-1}}\left[ xy-\sqrt{1-{{x}^{2}}}\sqrt{1-{{y}^{2}}} \right]=\pi -{{\cos }^{-1}}z $$

$$ \Rightarrow \left[ xy-\sqrt{1-{{x}^{2}}}\sqrt{1-{{y}^{2}}} \right]=\cos \left( \pi -{{\cos }^{-1}}z \right) $$

$$ \Rightarrow \left[ xy-\sqrt{1-{{x}^{2}}}\sqrt{1-{{y}^{2}}} \right]=\cos \left( \pi -{{\cos }^{-1}}z \right)\,\,\,\,\therefore \cos \left( \pi -\theta  \right)=-\cos \theta  $$

$$ \Rightarrow \left[ xy-\sqrt{1-{{x}^{2}}}\sqrt{1-{{y}^{2}}} \right]=-\cos {{\cos }^{-1}}z $$

$$ \Rightarrow \left[ xy-\sqrt{1-{{x}^{2}}}\sqrt{1-{{y}^{2}}} \right]=-z $$

$$ \Rightarrow xy+z=\sqrt{1-{{x}^{2}}}\sqrt{1-{{y}^{2}}} $$

On squaring both side and we get$$\begin{align}

$$ {{\left( xy+z \right)}^{2}}=\left( 1-{{x}^{2}} \right)\left( 1-{{y}^{2}} \right) $$

$$ \Rightarrow {{x}^{2}}{{y}^{2}}+{{z}^{2}}+2xyz=1-{{x}^{2}}-{{y}^{2}}+{{x}^{2}}{{y}^{2}} $$

$$ \Rightarrow {{z}^{2}}+2xyz=1-{{x}^{2}}-{{y}^{2}} $$

$$ \Rightarrow {{x}^{2}}+{{y}^{2}}+{{z}^{2}}+2xyz=1 $$

Hence proved.

We have,

$${{\sin }^{-1}}x=\dfrac{\pi }{5}$$

We know that,

$$ {{\sin }^{-1}}x+{{\cos }^{-1}}x=\dfrac{\pi }{2} $$

$$ {{\cos }^{-1}}x=\dfrac{\pi }{2}-{{\sin }^{-1}}x $$

Now,

$$ {{\cos }^{-1}}x=\dfrac{\pi }{2}-\dfrac{\pi }{5} $$

$$ =\dfrac{5\pi -2\pi }{10} $$

$$ =\dfrac{3\pi }{10} $$

$$tan^{-1}\left ( \dfrac{cos x}{1+sinx} \right )=tan^{-1}\left ( \dfrac{cos^2\dfrac{x}{2}-sin^2\dfrac{x}{2}}{cos^2\dfrac{x}{2}+sin^2\dfrac{x}{2}+2.cos\dfrac{x}{2}.sin\dfrac{x}{2}} \right )$$
$$=tan^{-1}\left [ \dfrac{\left ( cos\dfrac{x}{2}-sin\dfrac{x}{2} \right )\left (  cos\dfrac{x}{2}+sin\dfrac{x}{2}\right )}{\left ( cos\dfrac{x}{2}+\sin\dfrac{x}{2} \right )^2} \right ]$$
$$=tan^{-1} \left ( \dfrac{cos\dfrac{x}{2}-sin\dfrac{x}{2}}{cos\dfrac{x}{2}+sin\dfrac{x}{2}} \right )=tan^{-1}\left ( \dfrac{\dfrac{cos\dfrac{x}{2}}{cos\dfrac{x}{2}}-\dfrac{sin\dfrac{x}{2}}{cos\dfrac{x}{2}}}{\dfrac{cos\dfrac{x}{2}}{cos\dfrac{x}{2}}+\dfrac{sin\dfrac{x}{2}}{cos\dfrac{x}{2}}} \right )$$
$$=tan^{-1}\left ( \dfrac{1-tan\dfrac{x}{2}}{1+tan\dfrac{x}{2}} \right )=tan^{-1}\left ( \dfrac{tan\dfrac{\pi}{4}-tan\dfrac{x}{2}}{1+tan\dfrac{\pi}{2}tan\dfrac{x}{2}} \right )$$
$$=tan^{-1}\left [ tan\left ( \dfrac{\pi}{4}-\dfrac{x}{2} \right ) \right ]$$
$$=\dfrac{\pi}{4}-\dfrac{x}{2}$$

Let $$sin^{-1}\left ( \dfrac{8}{17} \right )=\alpha$$ and $$sin^{-1}\left ( \dfrac{3}{5} \right )=\beta$$
$$sin\alpha =\dfrac{8}{17}$$ and $$sin\beta=\dfrac{3}{5}$$
$$\Rightarrow cos\alpha=\sqrt{1-sin^2\alpha}$$ and $$cos\beta=\sqrt{1-sin^2\beta}$$
$$\Rightarrow cos\alpha=\sqrt{1-\dfrac{64}{289}}$$ and $$cos\beta=\sqrt{1-\dfrac{9}{25}}$$
$$\Rightarrow cos\alpha=\sqrt{\dfrac{289-64}{289}}$$ and $$cos\beta=\sqrt{\dfrac{25-9}{25}}$$
$$\Rightarrow cos\alpha=\sqrt{\dfrac{225}{289}}$$ and $$cos\beta=\sqrt{\dfrac{16}{25}}$$
$$\Rightarrow cos\alpha=\dfrac{15}{17}$$ and $$cos\beta=\dfrac{4}{5}$$
Now, $$cos(\alpha +\beta)=cos\alpha.cos\beta-sin \alpha. sin\beta$$
$$\Rightarrow cos(\alpha +\beta)=\dfrac{15}{17}\times \dfrac{4}{5}-\dfrac{8}{17}\times\dfrac{3}{5}$$
$$\Rightarrow cos(\alpha +\beta)= \dfrac{60}{85}-\dfrac{24}{85}$$
$$\Rightarrow cos(\alpha +\beta)= \dfrac{36}{85}$$
$$\Rightarrow \alpha +\beta=cos^{-1} \dfrac{36}{85}$$
Therefore, $$\Rightarrow sin^{-1}\dfrac{8}{17}+sin^{-1}\dfrac{3}{5}=cos^{-1}\dfrac{36}{85}$$
Hence proved.

Applying the trigonometric identities in the LHS of $$2{\sin ^{ - 1}}\left( {\dfrac{5}{{13}}} \right) = {\cos ^{ - 1}}\left( {\dfrac{{119}}{{169}}} \right)$$.

$$2{\sin ^{ - 1}}\left( {\dfrac{5}{{13}}} \right) = {\sin ^{ - 1}}\left( {2\left( {\dfrac{5}{{13}}} \right)\sqrt {1 - {{\left( {\dfrac{5}{{13}}} \right)}^2}} } \right)$$

$$ = {\sin ^{ - 1}}\left( {\dfrac{{10}}{{13}}\sqrt {1 - \dfrac{{25}}{{169}}} } \right)$$

$$ = {\sin ^{ - 1}}\left( {\dfrac{{10}}{{13}}\sqrt {\dfrac{{144}}{{169}}} } \right)$$

$$ = {\sin ^{ - 1}}\left( {\left( {\dfrac{{10}}{{13}}} \right)\left( {\dfrac{{12}}{{13}}} \right)} \right)$$

$$ = {\sin ^{ - 1}}\left( {\dfrac{{120}}{{169}}} \right)$$

Let $${\sin ^{ - 1}}\left( {\dfrac{{120}}{{169}}} \right) = \theta $$,

$$\dfrac{{120}}{{169}} = \sin \theta $$

$$\cos \theta  = \dfrac{{119}}{{169}}$$

$$\theta  = {\cos ^{ - 1}}\left( {\dfrac{{119}}{{169}}} \right)$$

$${\sin ^{ - 1}}\left( {\dfrac{{120}}{{169}}} \right) = {\cos ^{ - 1}}\left( {\dfrac{{119}}{{169}}} \right)$$

Therefore, $$2{\sin ^{ - 1}}\left( {\dfrac{5}{{13}}} \right) = {\cos ^{ - 1}}\left( {\dfrac{{119}}{{169}}} \right)$$.

Hence proved.

$${\sin ^{ - 1}}\left( {\sin \dfrac{{3\pi }}{5}} \right)$$

$$ = {\sin ^{ - 1}}\left( {\sin \left( {\pi  - \dfrac{{2\pi }}{5}} \right)} \right)$$

$$ = {\sin ^{ - 1}}\left( {\sin \left( {\dfrac{{2\pi }}{5}} \right)} \right)$$

$$ = \dfrac{{2\pi }}{5}$$

$$\sqrt{1+sinx}=\sqrt{sin^2(\frac{x}{2})+cos^2(\frac{x}{2})+2sin^2(\frac{x}{2})cos^2(\frac{x}{2})}\\=\>\sqrt{{(sin(\frac{x}{2})+cos(\frac{x}{2}))^2}}=(sin(\frac{x}{2})+cos(\frac{x}{2})\\\sqrt{1-sinx}=\sqrt{sin^2(\frac{x}{2})+cos^2(\frac{x}{2})-2sin^2(\frac{x}{2})cos^2(\frac{x}{2})}\\=\sqrt{{(cos(\frac{x}{2})-sin(\frac{x}{2}))^2}}=\>(cos(\frac{x}{2})-sin(\frac{x}{2}))\\\>\therefore\>cot^{-1}((\frac{(cos(\frac{x}{2})+sin(\frac{x}{2})+cos(\frac{x}{2})-sin(\frac{x}{2})}{(cos(\frac{x}{2})+sin(\frac{x}{2})-(cos(\frac{x}{2})+sin(\frac{x}{2})}))\\=cot^{-1}(cot(\frac{x}{2}))\\but\>this\>valid\>only\>for\>x\in\>(0,(\frac{\pi}{2}))$$