Inverse Trigonometric Functions
Let $S$ be the set of all solutions of the equation $\cos ^{-1}(2 x)-2 \cos ^{-1}\left(\sqrt{1-x^{2}}\right)=\pi, x \in\left[-\frac{1}{2}, \frac{1}{2}\right]$. Then $\sum_\limits{x \in S} 2 \sin ^{-1}\left(x^{2}-1\right)$ is equal to :
If $\alpha x^{2}+\beta x+\sin ^{-1}\left(x^{2}-6 x+10\right)+\cos ^{-1}\left(x^{2}-6 x+10\right)=0$ and $\alpha-\beta=b-a$, then $\alpha$ is equal to :
If ${\sin ^{ - 1}}{\alpha \over {17}} + {\cos ^{ - 1}}{4 \over 5} - {\tan ^{ - 1}}{{77} \over {36}} = 0,0 < \alpha < 13$, then ${\sin ^{ - 1}}(\sin \alpha ) + {\cos ^{ - 1}}(\cos \alpha )$ is equal to :
${\tan ^{ - 1}}\left( {{{1 + \sqrt 3 } \over {3 + \sqrt 3 }}} \right) + {\sec ^{ - 1}}\left( {\sqrt {{{8 + 4\sqrt 3 } \over {6 + 3\sqrt 3 }}} } \right)$ is equal to :
For $x \in(-1,1]$, the number of solutions of the equation $\sin ^{-1} x=2 \tan ^{-1} x$ is equal to __________.
Explanation:
We're given the equation $\sin^{-1}x = 2\tan^{-1}x$ for $x$ in the interval $(-1, 1]$. We want to find the number of solutions.
Step 1: Apply the sine and tangent functions to both sides :
We can rewrite the equation by applying the sine function to both sides :
$\sin(\sin^{-1}x) = \sin(2\tan^{-1}x).$
This simplifies to:
$x = \sin(2\tan^{-1}x).$
Step 2: Use the double-angle identity for sine :
Recall that $\sin(2y) = 2\sin(y)\cos(y)$. Applying this identity to the right-hand side gives :
$x = 2\sin(\tan^{-1}x)\cos(\tan^{-1}x).$
Step 3: Use the identities for sine and cosine of an inverse tangent :
Recall that $\sin(\tan^{-1}x) = \frac{x}{\sqrt{1 + x^2}}$ and $\cos(\tan^{-1}x) = \frac{1}{\sqrt{1 + x^2}}$. Substituting these into the equation gives :
$x = 2 \cdot \frac{x}{\sqrt{1 + x^2}} \cdot \frac{1}{\sqrt{1 + x^2}}.$
This simplifies to :
$x = \frac{2x}{1 + x^2}.$
Step 4: Solve for $x$ :
We have :
$x = \frac{2x}{1 + x^2}.$
Cross-multiplying gives :
$x(1 + x^2) = 2x.$
This simplifies to :
$x^3 + x - 2x = 0.$
Rearranging terms gives :
$x^3 - x = 0.$
This factors to:
$x(x^2 - 1) = 0.$
Setting each factor equal to zero gives the solutions $x = 0$, $x = -1$, and $x = 1$.
However, we are given that $x \in (-1, 1]$. Therefore, the only solutions in this interval are $x = 0$ and $x = 1$.
So there are 2 solutions to the equation $\sin ^{-1} x=2 \tan ^{-1} x$ in the interval $x \in(-1,1]$.
If $S=\left\{x \in \mathbb{R}: \sin ^{-1}\left(\frac{x+1}{\sqrt{x^{2}+2 x+2}}\right)-\sin ^{-1}\left(\frac{x}{\sqrt{x^{2}+1}}\right)=\frac{\pi}{4}\right\}$, then $\sum_\limits{x \in s}\left(\sin \left(\left(x^{2}+x+5\right) \frac{\pi}{2}\right)-\cos \left(\left(x^{2}+x+5\right) \pi\right)\right)$ is equal to ____________.
Explanation:
$ \sin^{-1}\left(\frac{x+1}{\sqrt{x^{2}+2 x+2}}\right)-\sin^{-1}\left(\frac{x}{\sqrt{x^{2}+1}}\right)=\frac{\pi}{4}. $
Let's denote:
$A = \sin^{-1}\left(\frac{x+1}{\sqrt{x^{2}+2 x+2}}\right),$
$B = \sin^{-1}\left(\frac{x}{\sqrt{x^{2}+1}}\right).$
So, we have the equation $A - B = \frac{\pi}{4}$.
We can also write this as $A = B + \frac{\pi}{4}$.
This gives us
$\sin(A) = \sin\left(B + \frac{\pi}{4}\right).$
We can use the identity $\sin(a + b) = \sin a \cos b + \cos a \sin b$ and rewrite this equation as:
$\frac{x+1}{\sqrt{(x+1)^2+1}} = \frac{x}{\sqrt{x^2+1}} \cos\left(\frac{\pi}{4}\right) + \sqrt{1-\left(\frac{x}{\sqrt{x^2+1}}\right)^2} \sin\left(\frac{\pi}{4}\right).$
After simplifying, we get:
$\frac{x+1}{\sqrt{x^2 + 2x + 2}} = \frac{1}{\sqrt{2}}\left(\frac{x}{\sqrt{x^2 + 1}} + \sqrt{1 - \frac{x^2}{x^2 + 1}}\right).$
Let's square both sides to remove the square roots:
On the left side, squaring gives:
$\left(\frac{x+1}{\sqrt{x^2 + 2x + 2}}\right)^2 = \frac{(x+1)^2}{x^2 + 2x + 2}.$
On the right side, squaring gives:
$\left(\frac{1}{\sqrt{2}}\left(\frac{x}{\sqrt{x^2 + 1}} + \sqrt{1 - \frac{x^2}{x^2 + 1}}\right)\right)^2 = \frac{1}{2}\left(\frac{x^2}{x^2+1} + 2\frac{x}{\sqrt{x^2+1}}\sqrt{1-\frac{x^2}{x^2+1}} + 1 - \frac{x^2}{x^2+1}\right).$
$ \therefore $ $\frac{(x+1)^2}{x^2 + 2x + 2}$ = $\frac{1}{2}\left(\frac{x^2}{x^2+1} + 2\frac{x}{\sqrt{x^2+1}}\sqrt{1-\frac{x^2}{x^2+1}} + 1 - \frac{x^2}{x^2+1}\right)$
$ \Rightarrow $ $\frac{(x+1)^2}{x^2 + 2x + 2}$ = ${1 \over 2}\left( {2 \times {x \over {\sqrt {{x^2} + 1} }}\sqrt {{{{x^2} + 1 - {x^2}} \over {{x^2} + 1}}} + 1} \right)$
$ \Rightarrow $ $\frac{(x+1)^2}{x^2 + 2x + 2}$ = ${1 \over 2}\left( {2 \times {x \over {\sqrt {{x^2} + 1} }}{1 \over {\sqrt {{x^2} + 1} }} + 1} \right)$
$ \Rightarrow $ $\frac{(x+1)^2}{x^2 + 2x + 2}$ = ${1 \over 2}\left( {2 \times {x \over {{x^2} + 1}} + 1} \right)$
$ \Rightarrow $ $\frac{(x+1)^2}{x^2 + 2x + 2}$ = ${1 \over 2}\left( {{{2x + {x^2} + 1} \over {{x^2} + 1}}} \right)$
$ \Rightarrow $ $\frac{(x+1)^2}{x^2 + 2x + 2}$ = ${1 \over 2}\left( {{{{{\left( {x + 1} \right)}^2}} \over {{x^2} + 1}}} \right)$
$ \begin{aligned} & \Rightarrow \frac{x+1}{\sqrt{x^2+2x+2}}=\frac{x+1}{\sqrt{2} \sqrt{x^2+1}} \\\\ & \Rightarrow x=-1 \text { OR } \sqrt{x^2+2x+2}=\sqrt{2} \cdot \sqrt{x^2+1} \\\\ & \Rightarrow x=0, x=2 \text { (Rejected) } \\\\ & S=\{0,-1\} \end{aligned} $
$ \begin{aligned} & \sum_{x \in R}\left(\sin \left(\left(x^2+x+5\right) \frac{\pi}{2}\right)-\cos \left(\left(x^2+x+5\right) \pi\right)\right) \\\\ & =\left[\sin \left(\frac{5 \pi}{2}\right)-\cos (5 \pi)\right]+\left[\sin \left(\frac{5 \pi}{2}\right)-\cos (5 \pi)\right] \\\\ & = (1 -(-1)) + (1 -(-1))\\\\ & = 2 + 2 \\\\ & = 4 \end{aligned} $
If the domain of the function $f(x)=\sec ^{-1}\left(\frac{2 x}{5 x+3}\right)$ is $[\alpha, \beta) \mathrm{U}(\gamma, \delta]$, then $|3 \alpha+10(\beta+\gamma)+21 \delta|$ is equal to _________.
Explanation:
Since, the domain for $\sec ^{-1} x$ is $|x| \geq 1$
Therefore $\left|\frac{2 x}{5 x+3}\right| \geq 1$
$ \begin{aligned} & \frac{2 x}{5 x+3} \leq-1 \text { or } \frac{2 x}{5 x+3} \geq 1 \\\\ & \frac{2 x+5 x+3}{5 x+3} \leq 0 \text { or } \frac{2 x-5 x-3}{5 x+3} \geq 0 \\\\ & \frac{7 x+3}{5 x+3} \leq 0 \text { or } \frac{-3(x+1)}{5 x+3} \geq 0 \end{aligned} $
Case I : $7 x+3 \leq 0$ and $5 x+3>0$
$ \begin{array}{rlrl} & x \leq-\frac{3}{7} \text { or } x > -\frac{3}{5} \\\\ & \Rightarrow -\frac{3}{5} < x \leq-\frac{3}{7} \end{array} $
Case II : $7 x+3 \geq 0$ and $5 x+3<0$
$ x \geq-\frac{3}{7} \text { and } x<-\frac{3}{5} $
Which is not possible
Case III : $x+1 \geq 0$ and $5 x+3<0$
$ \begin{aligned} & x \geq-1 \text { and } x<-\frac{3}{5} \\\\ & \Rightarrow -1 \leq x<-\frac{3}{5} \end{aligned} $
Case IV : $x+1 \leq 0$ and $5 x+3 \geq 0$
$ x \leq-1 \text { and } x \geq-\frac{3}{5} $
Which is not possible
$\therefore$ Domain is $\left[-1,-\frac{3}{5}\right) \cup\left(-\frac{3}{5},-\frac{3}{7}\right]$
$ \therefore \alpha=-1, \beta=-\frac{3}{5}, \gamma=-\frac{3}{5}, \delta=-\frac{3}{7} $
$ \therefore $ $ |3 \alpha+10(\beta+\gamma)+21 \delta| =\left|-3+10\left(-\frac{3}{5}-\frac{3}{5}\right)+21\left(-\frac{3}{7}\right)\right|=24 $
If the sum of all the solutions of ${\tan ^{ - 1}}\left( {{{2x} \over {1 - {x^2}}}} \right) + {\cot ^{ - 1}}\left( {{{1 - {x^2}} \over {2x}}} \right) = {\pi \over 3}, - 1 < x < 1,x \ne 0$, is $\alpha - {4 \over {\sqrt 3 }}$, then $\alpha$ is equal to _____________.
Explanation:
$-1 < x < 0$
$\tan ^{-1}\left(\frac{2 x}{1-x^{2}}\right)+\pi+\tan ^{-1}\left(\frac{2 x}{1-x^{2}}\right)=\frac{\pi}{3}$
$\tan ^{-1} \frac{2 x}{1-x^{2}}=\frac{-\pi}{3}$
$2 \tan ^{-1} x=\frac{-\pi}{3}$
$\tan ^{-1} x=\frac{-\pi}{6}$
$x=\frac{-1}{\sqrt{3}}$
Case-II
$0 < x < 1$
$\tan ^{-1} \frac{2 x}{1-x^{2}}+\tan ^{-1} \frac{2 x}{1-x^{2}}=\frac{\pi}{3}$
$ \begin{aligned} & \tan ^{-1} \frac{2 x}{1-x^{2}}=\frac{\pi}{6} \\\\ & 2 \tan ^{-1} x=\frac{\pi}{6} \\\\ & \tan ^{-1} x=\frac{\pi}{12} \\\\ & x=2-\sqrt{3} \\\\ & \text { Sum }=\frac{-1}{\sqrt{3}}+2-\sqrt{3}=2-\frac{4}{\sqrt{3}} \\\\ & \Rightarrow \alpha=2 \end{aligned} $
Then $\tan ^{-1}\left(\frac{1}{1+a_{1} a_{2}}\right)+\tan ^{-1}\left(\frac{1}{1+a_{2} a_{3}}\right)+\ldots . .+\tan ^{-1}\left(\frac{1}{1+a_{2021} a_{2022}}\right)$ is equal to :
$\tan ^{-1}\left(\frac{6 y}{9-y^2}\right)+\cot ^{-1}\left(\frac{9-y^2}{6 y}\right)=\frac{2 \pi}{3}$ for $0<|y|<3$, is equal to :
Explanation:
$\begin{aligned} & \Rightarrow \sqrt{2 \cos ^2 x}=\sqrt{2} \tan ^{-1} \tan x \\\\ & \Rightarrow \sqrt{2}|\cos x|=\sqrt{2} \tan ^{-1} \tan x \\\\ & \Rightarrow |\cos x|=\tan ^{-1} \tan x\end{aligned}$
Number of solution $=3$.
The domain of the function $f(x)=\sin ^{-1}\left(\frac{x^{2}-3 x+2}{x^{2}+2 x+7}\right)$ is :
The sum of the absolute maximum and absolute minimum values of the function $f(x)=\tan ^{-1}(\sin x-\cos x)$ in the interval $[0, \pi]$ is :
Considering only the principal values of the inverse trigonometric functions, the domain of the function $f(x)=\cos ^{-1}\left(\frac{x^{2}-4 x+2}{x^{2}+3}\right)$ is :
Considering the principal values of the inverse trigonometric functions, the sum of all the solutions of the equation $\cos ^{-1}(x)-2 \sin ^{-1}(x)=\cos ^{-1}(2 x)$ is equal to :
The domain of the function $f(x)=\sin ^{-1}\left[2 x^{2}-3\right]+\log _{2}\left(\log _{\frac{1}{2}}\left(x^{2}-5 x+5\right)\right)$, where [t] is the greatest integer function, is :
If $0 < x < {1 \over {\sqrt 2 }}$ and ${{{{\sin }^{ - 1}}x} \over \alpha } = {{{{\cos }^{ - 1}}x} \over \beta }$, then the value of $\sin \left( {{{2\pi \alpha } \over {\alpha + \beta }}} \right)$ is :
$\tan \left(2 \tan ^{-1} \frac{1}{5}+\sec ^{-1} \frac{\sqrt{5}}{2}+2 \tan ^{-1} \frac{1}{8}\right)$ is equal to :
Let m and M respectively be the minimum and the maximum values of $f(x) = {\sin ^{ - 1}}2x + \sin 2x + {\cos ^{ - 1}}2x + \cos 2x,\,x \in \left[ {0,{\pi \over 8}} \right]$. Then m + M is equal to :
Let $\alpha = \tan \left( {{{5\pi } \over {16}}\sin \left( {2{{\cos }^{ - 1}}\left( {{1 \over {\sqrt 5 }}} \right)} \right)} \right)$ and $\beta = \cos \left( {{{\sin }^{ - 1}}\left( {{4 \over 5}} \right) + {{\sec }^{ - 1}}\left( {{5 \over 3}} \right)} \right)$ where the inverse trigonometric functions take principal values. Then, the equation whose roots are $\alpha$ and $\beta$ is :
The domain of the function ${\cos ^{ - 1}}\left( {{{2{{\sin }^{ - 1}}\left( {{1 \over {4{x^2} - 1}}} \right)} \over \pi }} \right)$ is :
The value of $\cot \left( {\sum\limits_{n = 1}^{50} {{{\tan }^{ - 1}}\left( {{1 \over {1 + n + {n^2}}}} \right)} } \right)$ is :
${\sin ^1}\left( {\sin {{2\pi } \over 3}} \right) + {\cos ^{ - 1}}\left( {\cos {{7\pi } \over 6}} \right) + {\tan ^{ - 1}}\left( {\tan {{3\pi } \over 4}} \right)$ is equal to :
If the inverse trigonometric functions take principal values then
${\cos ^{ - 1}}\left( {{3 \over {10}}\cos \left( {{{\tan }^{ - 1}}\left( {{4 \over 3}} \right)} \right) + {2 \over 5}\sin \left( {{{\tan }^{ - 1}}\left( {{4 \over 3}} \right)} \right)} \right)$ is equal to :
The value of ${\tan ^{ - 1}}\left( {{{\cos \left( {{{15\pi } \over 4}} \right) - 1} \over {\sin \left( {{\pi \over 4}} \right)}}} \right)$ is equal to :
Let $x * y = {x^2} + {y^3}$ and $(x * 1) * 1 = x * (1 * 1)$.
Then a value of $2{\sin ^{ - 1}}\left( {{{{x^4} + {x^2} - 2} \over {{x^4} + {x^2} + 2}}} \right)$ is :
The set of all values of k for which
${({\tan ^{ - 1}}x)^3} + {({\cot ^{ - 1}}x)^3} = k{\pi ^3},\,x \in R$, is the interval :
The domain of the function
$f(x) = {{{{\cos }^{ - 1}}\left( {{{{x^2} - 5x + 6} \over {{x^2} - 9}}} \right)} \over {{{\log }_e}({x^2} - 3x + 2)}}$ is :
For $k \in \mathbb{R}$, let the solutions of the equation $\cos \left(\sin ^{-1}\left(x \cot \left(\tan ^{-1}\left(\cos \left(\sin ^{-1} x\right)\right)\right)\right)\right)=k, 0<|x|<\frac{1}{\sqrt{2}}$ be $\alpha$ and $\beta$, where the inverse trigonometric functions take only principal values. If the solutions of the equation $x^{2}-b x-5=0$ are $\frac{1}{\alpha^{2}}+\frac{1}{\beta^{2}}$ and $\frac{\alpha}{\beta}$, then $\frac{b}{k^{2}}$ is equal to ____________.
Explanation:
$\cos \left( {{{\sin }^{ - 1}}\left( {x\cot \left( {{{\tan }^{ - 1}}\left( {\cos \left( {{{\sin }^{ - 1}}} \right)} \right)} \right)} \right)} \right) = k$
$ \Rightarrow \cos \left( {{{\sin }^{ - 1}}\left( {x\cot \left( {{{\tan }^{ - 1}}\sqrt {1 - {x^2}} } \right)} \right)} \right) = k$
$ \Rightarrow \cos \left( {{{\sin }^{ - 1}}\left( {{x \over {\sqrt {1 - {x^2}} }}} \right)} \right) = k$
$ \Rightarrow {{\sqrt {1 - 2{x^2}} } \over {\sqrt {1 - {x^2}} }} = k$
$ \Rightarrow {{1 - 2{x^2}} \over {1 - {x^2}}} = {k^2}$
$ \Rightarrow 1 - 2{x^2} = {k^2} - {k^2}{x^2}$
$\therefore$
${1 \over {{\alpha ^2}}} + {1 \over {{\beta ^2}}} = 2\left( {{{{k^2} - 2} \over {{k^2} - 1}}} \right)$ ...... (1)
and ${\alpha \over \beta } = - 1$ ...... (2)
$\therefore$ $2\left( {{{{k^2} - 2} \over {{k^2} - 1}}} \right)( - 1) = - 5$
$ \Rightarrow {k^2} = {1 \over 3}$
and $b = S.R = 2\left( {{{{k^2} - 2} \over {{k^2} - 1}}} \right) - 1 = 4$
$\therefore$ ${b \over {{k^2}}} = {4 \over {{1 \over 3}}} = 12$
Let $x = \sin (2{\tan ^{ - 1}}\alpha )$ and $y = \sin \left( {{1 \over 2}{{\tan }^{ - 1}}{4 \over 3}} \right)$. If $S = \{ a \in R:{y^2} = 1 - x\} $, then $\sum\limits_{\alpha \in S}^{} {16{\alpha ^3}} $ is equal to _______________.
Explanation:
$\because$ $x = \sin \left( {2{{\tan }^{ - 1}}\alpha } \right) = {{2\alpha } \over {1 + {\alpha ^2}}}$ ...... (i)
and $y = \sin \left( {{1 \over 2}{{\tan }^{ - 1}}{4 \over 3}} \right) = \sin \left( {{{\sin }^{ - 1}}{1 \over {\sqrt 5 }}} \right) = {1 \over {\sqrt 5 }}$
Now, ${y^2} = 1 - x$
${1 \over 5} = 1 - {{2\alpha } \over {1 + {\alpha ^2}}}$
$ \Rightarrow 1 + {\alpha ^2} = 5 + 5{\alpha ^2} - 10\alpha $
$ \Rightarrow 2{\alpha ^2} - 5\alpha + 2 = 0$
$\therefore$ $\alpha = 2,{1 \over 2}$
$\therefore$ $\sum\limits_{\alpha \in S} {16{\alpha ^3} = 16 \times {2^3} + 16 \times {1 \over {{2^3}}} = 130} $
$50\tan \left( {3{{\tan }^{ - 1}}\left( {{1 \over 2}} \right) + 2{{\cos }^{ - 1}}\left( {{1 \over {\sqrt 5 }}} \right)} \right) + 4\sqrt 2 \tan \left( {{1 \over 2}{{\tan }^{ - 1}}(2\sqrt 2 )} \right)$ is equal to ____________.
Explanation:
$\Rightarrow 50 \tan \left(\pi+\tan ^{-1}\left(\frac{1}{2}\right)\right)+4 \sqrt{2} \tan \left(\frac{1}{2} \tan ^{-1} 2 \sqrt{2}\right)$
$\Rightarrow \quad 50\left(\frac{1}{2}\right)+4 \sqrt{2} \tan \alpha$
Where $2 \alpha=\tan ^{-1} 2 \sqrt{2}$
$\Rightarrow \frac{2 \tan \alpha}{1-\tan ^{2} \alpha}=2 \sqrt{2} \quad$.. (i)
$\Rightarrow \quad 2 \sqrt{2} \tan ^{2} \alpha+2 \tan \alpha-2 \sqrt{2}=0$
$\Rightarrow \quad 2 \sqrt{2} \tan ^{2} \alpha+4 \tan \alpha-2 \tan \alpha-2 \sqrt{2}=0$
$\Rightarrow(2 \sqrt{2} \tan \alpha-2)(\tan \alpha-\sqrt{2})=0$
$\Rightarrow \tan \alpha=\sqrt{2}$ or $\frac{1}{\sqrt{2}}$
$\Rightarrow \tan \alpha=\frac{1}{\sqrt{2}}$
$(\tan \alpha=\sqrt{2}$ doesn't satisfy (i))
$\Rightarrow \quad 25+4 \sqrt{2} \frac{1}{\sqrt{2}}=29$
$ \frac{3}{2} \cos ^{-1} \sqrt{\frac{2}{2+\pi^{2}}}+\frac{1}{4} \sin ^{-1} \frac{2 \sqrt{2} \pi}{2+\pi^{2}}+\tan ^{-1} \frac{\sqrt{2}}{\pi} $
is
Explanation:
Given, ${3 \over 2}{\cos ^{ - 1}}\sqrt {{2 \over {2 + {\pi ^2}}}} + {1 \over 4}{\sin ^{ - 1}}\left( {{{2\sqrt 2 \pi } \over {2 + {\pi ^2}}}} \right) + {\tan ^{ - 1}}\left( {{{\sqrt 2 } \over \pi }} \right)$
Let, ${\tan ^{ - 1}}\left( {{{\sqrt 2 } \over \pi }} \right) = \alpha $
$ \Rightarrow {{\sqrt 2 } \over \pi } = \tan \alpha $
We know, $\sin 2\alpha = {{2\tan \alpha } \over {1 + {{\tan }^2}\alpha }}$
$ = {{2\left( {{{\sqrt 2 } \over \pi }} \right)} \over {1 + {2 \over {{\pi ^2}}}}}$
$ = {{2\sqrt 2 \pi } \over {2 + {\pi ^2}}}$
$ \Rightarrow 2\alpha = {\sin ^{ - 1}}\left( {{{2\sqrt 2 \pi } \over {2 + {\pi ^2}}}} \right) = 2{\tan ^{ - 1}}\left( {{{\sqrt 2 } \over \pi }} \right)$
Now, let, ${\cos ^{ - 1}}\left( {\sqrt {{2 \over {2 + {\pi ^2}}}} } \right) = \beta $
$ \Rightarrow \sqrt {{2 \over {2 + {\pi ^2}}}} = \cos \beta $
$\therefore$ $\cot \beta = {{\sqrt 2 } \over \pi }$
$ \Rightarrow \beta = {\cot ^{ - 1}}\left( {{{\sqrt 2 } \over \pi }} \right)$
$\therefore$ ${\cos ^{ - 1}}\left( {\sqrt {{2 \over {2 + {\pi ^2}}}} } \right) = {\cot ^{ - 1}}\left( {{{\sqrt 2 } \over \pi }} \right)$
$\therefore$ ${3 \over 2}{\cos ^{ - 1}}\left( {\sqrt {{2 \over {2 + {\pi ^2}}}} } \right) + {1 \over 4}{\sin ^{ - 1}}\left( {{{2\sqrt 2 \pi } \over {2 + {\pi ^2}}}} \right) + {\tan ^{ - 1}}\left( {{{\sqrt 2 } \over \pi }} \right)$
$ = {3 \over 2}{\cot ^{ - 1}}\left( {{{\sqrt 2 } \over \pi }} \right) + {1 \over 4}\left( {2 + \tan \left( {{{\sqrt 2 } \over \pi }} \right)} \right) + {\tan ^{ - 1}}\left( {{{\sqrt 2 } \over \pi }} \right)$
$ = {3 \over 2}{\cot ^{ - 1}}\left( {{{\sqrt 2 } \over \pi }} \right) + {1 \over 2}{\tan ^{ - 1}}\left( {{{\sqrt 2 } \over \pi }} \right) + {\tan ^{ - 1}}\left( {{{\sqrt 2 } \over \pi }} \right)$
$ = {3 \over 2}{\cot ^{ - 1}}\left( {{{\sqrt 2 } \over \pi }} \right) + {3 \over 2}{\tan ^{ - 1}}\left( {{{\sqrt 2 } \over \pi }} \right)$
$ = {3 \over 2}\left( {{{\tan }^{ - 1}}\left( {{{\sqrt 2 } \over \pi }} \right) + {{\cot }^{ - 1}}\left( {{{\sqrt 2 } \over \pi }} \right)} \right)$
$ = {3 \over 2} \times {\pi \over 2}$
$ = {{3\pi } \over 4}$
$ = 2.36$
(The inverse trigonometric functions take the principal values)
$f(x) = {\sin ^{ - 1}}\left( {{{3{x^2} + x - 1} \over {{{(x - 1)}^2}}}} \right) + {\cos ^{ - 1}}\left( {{{x - 1} \over {x + 1}}} \right)$ is :
f(x) = tan$-$1 (sin x + cos x) in $\left[ {0,{\pi \over 2}} \right]$, then the value of tan(M $-$ m) is equal to :
${\sin ^{ - 1}}\left[ {{x^2} + {1 \over 3}} \right] + {\cos ^{ - 1}}\left[ {{x^2} - {2 \over 3}} \right] = {x^2}$, for x$\in$[$-$1, 1], and [x] denotes the greatest integer less than or equal to x, is :
tan$-$1(x + 1) + cot$-$1$\left( {{1 \over {x - 1}}} \right)$ = tan$-$1$\left( {{8 \over {31}}} \right)$ is :
${\sin ^{ - 1}}\left( {{{3x} \over 5}} \right) + {\sin ^{ - 1}}\left( {{{4x} \over 5}} \right) = {\sin ^{ - 1}}x$ is equal to :
$(a + b) - \left( {{{{a^2} + {b^2}} \over 2}} \right) + \left( {{{{a^3} + {b^3}} \over 3}} \right) - \left( {{{{a^4} + {b^4}} \over 4}} \right) + .....$ is :
then the value of $\cos \left( {{{\pi c} \over {a + b}}} \right)$ is :
$\tan ^{-1}(-2)-\tan ^{-1}(3)$ is equal to