Inverse Trigonometric Functions

Given :

$ k=\tan \left(\frac{\pi}{4}+\frac{1}{2} \cos ^{-1}\left(\frac{2}{3}\right)\right)+\tan \left(\frac{1}{2} \sin ^{-1}\left(\frac{2}{3}\right)\right) $

Let $\cos ^{-1} \frac{2}{3}=\theta$, so $\sin ^{-1} \frac{2}{3}=\frac{\pi}{2}-\theta$

$k=\tan \left(\frac{\pi}{4}+\frac{\theta}{2}\right)+\tan \left(\frac{\pi}{4}-\frac{\theta}{2}\right)$

$\Rightarrow $ $k=\frac{1+\tan \frac{\theta}{2}}{1-\tan \frac{\theta}{2}}+\frac{1-\tan \frac{\theta}{2}}{1+\tan \frac{\theta}{2}}$

$\Rightarrow $ $k=\frac{\left(1+\tan \frac{\theta}{2}\right)^2+\left(1-\tan \frac{\theta}{2}\right)^2}{\left(1-\tan \frac{\theta}{2}\right)\left(1+\tan \frac{\theta}{2}\right)}$

$\Rightarrow $ $k=\frac{1+\tan ^2 \frac{\theta}{2}+2 \tan \frac{\theta}{2}+1+\tan ^2 \frac{\theta}{2}-2 \tan \frac{\theta}{2}}{\left(1-\tan \frac{\theta}{2}\right)\left(1+\tan \frac{\theta}{2}\right)}$

$ k=\frac{2\left(1+\tan ^2 \frac{\theta}{2}\right)}{\left(1-\tan ^2 \frac{\theta}{2}\right)} $

Using the identity $\cos 2 x=\frac{1-\tan ^2 x}{1+\tan ^2 x}$ :

$k=\frac{2}{\cos \theta}$

$ \begin{aligned} & \text { Since } \theta=\cos ^{-1} \frac{2}{3} \Rightarrow \cos \theta=\frac{2}{3} \\ & k=\frac{2}{2 / 3}=3 \end{aligned} $

Solving the main equation :

$ \sin ^{-1}(k x-1)=\sin ^{-1} x-\cos ^{-1} x $

Put $k=3$ :

$\sin ^{-1}(3 x-1)=\sin ^{-1} x-\left(\frac{\pi}{2}-\sin ^{-1} x\right)$

$\Rightarrow $ $\sin ^{-1}(3 x-1)=2 \sin ^{-1} x-\frac{\pi}{2}$ ......(1)

Domain of $\sin ^{-1}(3 x-1) \Rightarrow 3 x-1 \in[-1,1] \Rightarrow x \in\left[0, \frac{2}{3}\right]$

Domain of $\sin ^{-1} x \Rightarrow x \in[-1,1]$

Intersection domain of Equation 1 is $x \in\left[0, \frac{2}{3}\right]$

Take sine of both sides of Equation 1:

$ 3 x-1=\sin \left(2 \sin ^{-1} x-\frac{\pi}{2}\right) $

$\Rightarrow $ $ 3 x-1=-\cos \left(2 \sin ^{-1} x\right) $

Using $\cos 2 A=1-2 \sin ^2 A$ :

$ 3 x-1=-\left(1-2\left(\sin \left(\sin ^{-1} x\right)\right)^2\right) $

$\Rightarrow $ $3 x-1=2 x^2-1$

$\Rightarrow $ $3 x=2 x^2 \Rightarrow x(2 x-3)=0$

$\Rightarrow $ $ x=0 \text { or } x=\frac{3}{2} $

$\frac{3}{2}$ is outside the domain.

So, $x=0$ is the only solution.

Number of solutions $=1$.

$ f(x)=\left(\sin^{-1}x\right)^2+\left(\cos^{-1}x\right)^2,\qquad x\in\left[-\frac{\sqrt3}{2},\frac1{\sqrt2}\right] $

For $x\in[-1,1]$,

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

Let $a=\sin^{-1}x$. Then $\cos^{-1}x=\frac{\pi}{2}-a$, and

$ f= a^2+\left(\frac{\pi}{2}-a\right)^2 =2a^2-\pi a+\frac{\pi^2}{4}. $

This is a quadratic in $a$ opening upward, so its maximum on the allowed interval occurs at an endpoint of $a$.

Now,

$ x=-\frac{\sqrt3}{2}\Rightarrow a=-\frac{\pi}{3},\qquad x=\frac1{\sqrt2}\Rightarrow a=\frac{\pi}{4}. $

Evaluate $f$:

• At $a=\frac{\pi}{4}$: this is the vertex, giving the minimum.

• At $a=-\frac{\pi}{3}$:

$ f=\left(-\frac{\pi}{3}\right)^2+\left(\frac{5\pi}{6}\right)^2 =\frac{\pi^2}{9}+\frac{25\pi^2}{36} =\frac{29\pi^2}{36}. $

Hence the maximum value is

$ \frac{m}{n}\pi^2=\frac{29}{36}\pi^2, $

with $\gcd(29,36)=1$.

$ \boxed{m+n=65} $

$\begin{aligned} & y=\cos \left(\frac{\pi}{3}+\cos ^{-1} \frac{x}{2}\right) \\ & =\cos \left(\frac{\pi}{3}\right) \cos \left(\cos ^{-1}\left(\frac{x}{2}\right)\right)-\sin \left(\frac{\pi}{3 .}\right) \sin \left(\cos ^{-1}\left(\frac{x}{2}\right)\right) \\ & =\frac{1}{2} \cdot \frac{x}{2}-\frac{\sqrt{3}}{2} \cdot \sqrt{1-\frac{x^2}{4}} \\ & \Rightarrow 4 y=x-\sqrt{3} \sqrt{4-x^2} \\ & \Rightarrow(4 y-x)^2=3\left(4-x^2\right) \\ & \Rightarrow 16 y^2+x^2-8 x y=12-3 x^2 \\ & x^2+4 y^2-2 x y=3 \\ & (x-y)^2+3 y^2=3 \end{aligned}$

$\begin{aligned} & \cos ^{-1} x=\pi+\sin ^{-1} x+\sin ^{-1}(2 x+1) \\ & 2 \cos ^{-1} x-\sin ^{-1}(2 x+1)=\frac{3 \pi}{2} \\ & 2 \alpha-\beta=\frac{3 \pi}{2} \text { where } \cos ^{-1} x=\alpha, \sin ^{-1}(2 x+1)=\beta \\ & 2 \alpha=\frac{3 \pi}{2}+\beta \\ & \cos 2 \alpha=\sin \beta \\ & \begin{array}{l} 2 \cos ^2 \alpha-1=\sin \beta \\ 2 x^2-1=2 x+1 \\ x^2-x-1=0 \\ \Rightarrow x=\frac{1-\sqrt{5}}{2},\left\{x=\frac{1+\sqrt{5}}{2} \text { rejected }\right\} \\ \therefore 4 x^2-4 x=4 \\ (2 x-1)^2=5 \end{array} \end{aligned}$

$\begin{aligned} & \operatorname{If}\left(\tan \left(\tan ^{-1}(\alpha)\right)+1\left(\cot \left(\cot ^{-1} \beta\right)\right)^2=36\right. \\ & \alpha^2+\beta^2=34 \\ & \alpha \beta=15 \\ & \alpha=3, \beta=5 \\ & \therefore \alpha^2+\beta=9+5=14 \end{aligned}$

$\begin{aligned} & 2 \sin ^{-1} x+3 \cos ^{-1} x=\frac{2 \pi}{5} \\ & \frac{\pi}{2}+\cos ^{-1} x=\frac{2 \pi}{5} \\ & \cos ^{-1} x=\frac{2 \pi}{5}-\frac{\pi}{2} \\ & \cos ^{-1} x=\frac{-\pi}{10} \end{aligned}$

Which is not possible as $\cos ^{-1} x \in[0, \pi]$

$\therefore \quad$ No solution

For $ n \in \mathbb{N} $, if $ \cot^{-1} 3 + \cot^{-1} 4 + \cot^{-1} 5 + \cot^{-1} n = \frac{\pi}{4} $, then $ n $ is equal to .

Given the equation:

$ \cot^{-1} 3 + \cot^{-1} 4 + \cot^{-1} 5 + \cot^{-1}(n) = \frac{\pi}{4} $

we can use the identity for the sum of inverse cotangents. Starting with the first two terms:

$ \cot^{-1} 3 + \cot^{-1} 4 = \cot^{-1}\left(\frac{3 \times 4 - 1}{3 + 4}\right) = \cot^{-1}\left(\frac{11}{7}\right) $

Now, adding the third term:

$ \cot^{-1}\left(\frac{11}{7}\right) + \cot^{-1}(5) $

we apply the identity again:

$ \cot^{-1}\left(\frac{11}{7}\right) + \cot^{-1}\left(\frac{n \times 5 - 1}{5 + n}\right) = \frac{\pi}{4} $

Rewriting this to isolate the sum of the terms, we proceed as follows:

$ \cot^{-1}\left( \frac{\left(\frac{11}{7} \times \frac{5n-1}{5+n} - 1\right)}{\left(\frac{11}{7} + \frac{5n-1}{5+n} \right)} \right) = \frac{\pi}{4} $

This simplifies to:

$ \frac{11}{7} \left(\frac{5n-1}{5+n}\right) - 1 = \frac{11}{7} + \frac{5n-1}{5+n} $

Solving the equation:

$ \frac{55n - 11}{5 + n} - 1 = \frac{11}{7} + \frac{5n - 1}{5 + n} $

Further simplification yields:

$ 55n - 11 - 35 - 7n = 55 + 11n + 35n - 7 $

Bringing the terms together, we get:

$ 48n - 46 = 48 $

Therefore:

$ 2n = 94 $

So finally:

$ n = 47 $

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 $x \in(-1,1]$.

Given equation is

$ \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} $

Given that $f(x)=\sec ^{-1}\left(\frac{2 x}{5 x+3}\right)$

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 $

Case-I

$-1 < x < 0$

$\tan ^{-1}\left(\frac{2 x}{1-x^{2}}\right)+\pi+\tan ^{-1}\left(\frac{2 x}{1-x^{2}}\right)=\frac{\pi}{3}$

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

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

$\tan ^{-1} x=\frac{-\pi}{6}$

$x=\frac{-1}{\sqrt{3}}$

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} $

$\cos \left( {{{\sin }^{ - 1}}\left( {x\cot \left( {{{\tan }^{ - 1}}\left( {\cos \left( {{{\sin }^{ - 1}}} \right)} \right)} \right)} \right)} \right) = k$

$ \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 $b = S.R = 2\left( {{{{k^2} - 2} \over {{k^2} - 1}}} \right) - 1 = 4$

$\therefore$ ${b \over {{k^2}}} = {4 \over {{1 \over 3}}} = 12$

$\because$ $x = \sin \left( {2{{\tan }^{ - 1}}\alpha } \right) = {{2\alpha } \over {1 + {\alpha ^2}}}$ ...... (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} $

$50 \tan \left(\tan ^{-1} \frac{1}{2}+2 \tan ^{-1}\left(\frac{1}{2}\right)+2 \tan ^{-1}(2)\right)$ $ +4 \sqrt{2} \tan \left(\frac{\tan ^{-1}}{2}(2 \sqrt{2})\right) $

$\Rightarrow 50 \tan \left(\pi+\tan ^{-1}\left(\frac{1}{2}\right)\right)+4 \sqrt{2} \tan \left(\frac{1}{2} \tan ^{-1} 2 \sqrt{2}\right)$

$\Rightarrow \quad 50\left(\frac{1}{2}\right)+4 \sqrt{2} \tan \alpha$

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))

$\Rightarrow \quad 25+4 \sqrt{2} \frac{1}{\sqrt{2}}=29$