Inverse Trigonometric Functions
Considering the principal values of inverse trigonometric functions, the value of the expression
$\tan \left( 2 \sin^{-1}\left( \frac{2}{\sqrt{13}} \right) - 2 \cos^{-1}\left( \frac{3}{\sqrt{10}} \right) \right)$
is equal to :
$ \frac{33}{56} $
$ -\frac{33}{56} $
$ -\frac{16}{63} $
$ \frac{16}{63} $
If the domain of the function $f(x)=\sin ^{-1}\left(\frac{1}{x^2-2 x-2}\right)$, is $(-\infty, \alpha] \cup[\beta, \gamma] \cup[\delta, \infty)$, then $\alpha+\beta+\gamma+\delta$ is equal to
4
2
5
3
The number of solutions of $\tan ^{-1} 4 x+\tan ^{-1} 6 x=\frac{\pi}{6}$, where $-\frac{1}{2 \sqrt{6}} < x < \frac{1}{2 \sqrt{6}}$, is equal to :
2
0
3
1
If the domain of the function $f(x)=\cos ^{-1}\left(\frac{2 x-5}{11-3 x}\right)+\sin ^{-1}\left(2 x^2-3 x+1\right)$ is the interval $[\alpha, \beta]$, then $\alpha+2 \beta$ is equal to :
5
2
3
1
If $k=\tan \left(\frac{\pi}{4}+\frac{1}{2} \cos ^{-1}\left(\frac{2}{3}\right)\right)+\tan \left(\frac{1}{2} \sin ^{-1}\left(\frac{2}{3}\right)\right)$, then
the number of solutions of the equation $\sin ^{-1}(k x-1)=\sin ^{-1} x-\cos ^{-1} x$ is $\_\_\_\_$.
Explanation:
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 $=1$.
Explanation:
$ 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} $
Let $\alpha=3 \sin ^{-1}\left(\frac{6}{11}\right)$ and $\beta=3 \cos ^{-1}\left(\frac{4}{9}\right)$, where inverse trigonometric functions take only the principal values.
Given below are two statements :
Statement I : $\quad \cos (\alpha+\beta)>0$.
Statement II : $\quad \cos (\alpha)<0$.
In the light of the above statements, choose the correct answer from the options given below :
Both Statement I and Statement II are true
Both Statement I and Statement II are false
Statement I is true but Statement II is false
$ \text { Statement I is false but Statement II is true } $
If $\sin \left(\tan ^{-1}(x \sqrt{2})\right)=\cot \left(\sin ^{-1} \sqrt{1-x^2}\right), x \in(0,1)$, then the value of $x$ is:
$\frac{1}{2}$
${\frac{1}{3}}$
Let $0<\alpha<1, \beta=\frac{1}{3 \alpha}$ and $\tan ^{-1}(1-\alpha)+\tan ^{-1}(1-\beta)=\frac{\pi}{4}$. Then $6(\alpha+\beta)$ is equal to:
If $\frac{\pi}{4}+\sum\limits_{p=1}^{11} \tan ^{-1}\left(\frac{2^{p-1}}{1+2^{2 p-1}}\right)=\alpha$, then $\tan \alpha$ is equal to $\_\_\_\_$ .
Explanation:
Use the identity
$ \tan^{-1}\left(\frac{x-y}{1+xy}\right)=\tan^{-1}x-\tan^{-1}y $
when $xy>-1$.
Now observe the given term:
$ \frac{2^{p-1}}{1+2^{2p-1}} $
We want to write it in the form
$ \frac{x-y}{1+xy} $
Take
$ x=2^p,\qquad y=2^{p-1} $
Then
$ x-y=2^p-2^{p-1}=2^{p-1} $
and
$ xy=2^p\cdot 2^{p-1}=2^{2p-1} $
So,
$ \frac{2^{p-1}}{1+2^{2p-1}} = \frac{2^p-2^{p-1}}{1+2^p\cdot 2^{p-1}} $
Hence,
$ \tan^{-1}\left(\frac{2^{p-1}}{1+2^{2p-1}}\right) = \tan^{-1}(2^p)-\tan^{-1}(2^{p-1}) $
Therefore,
$ \sum_{p=1}^{11}\tan^{-1}\left(\frac{2^{p-1}}{1+2^{2p-1}}\right) = \sum_{p=1}^{11}\left[\tan^{-1}(2^p)-\tan^{-1}(2^{p-1})\right] $
This is a telescoping sum:
$ = \left(\tan^{-1}2-\tan^{-1}1\right) +\left(\tan^{-1}4-\tan^{-1}2\right) +\cdots+ \left(\tan^{-1}2^{11}-\tan^{-1}2^{10}\right) $
So all middle terms cancel, and we get
$ \sum_{p=1}^{11}\tan^{-1}\left(\frac{2^{p-1}}{1+2^{2p-1}}\right) = \tan^{-1}(2^{11})-\tan^{-1}(1) $
Since
$ \tan^{-1}(1)=\frac{\pi}{4} $
given
$ \alpha=\frac{\pi}{4}+\sum_{p=1}^{11}\tan^{-1}\left(\frac{2^{p-1}}{1+2^{2p-1}}\right) $
becomes
$ \alpha=\frac{\pi}{4}+\tan^{-1}(2^{11})-\frac{\pi}{4} =\tan^{-1}(2^{11}) $
Thus,
$ \tan\alpha=\tan\left(\tan^{-1}(2^{11})\right)=2^{11}=2048 $
So, the answer is
$ \boxed{2048} $
The value of $ \cot^{-1} \left( \frac{\sqrt{1 + \tan^2(2)} - 1}{\tan(2)} \right) - \cot^{-1} \left( \frac{\sqrt{1 + \tan^2\left(\frac{1}{2}\right)} + 1}{\tan\left(\frac{1}{2}\right)} \right) $ is equal to
$ \pi - \frac{3}{2} $
$ \pi + \frac{5}{2} $
$ \pi - \frac{5}{4} $
$ \pi + \frac{3}{2} $
The sum of the infinite series $\cot ^{-1}\left(\frac{7}{4}\right)+\cot ^{-1}\left(\frac{19}{4}\right)+\cot ^{-1}\left(\frac{39}{4}\right)+\cot ^{-1}\left(\frac{67}{4}\right)+\ldots$. is :
Considering the principal values of the inverse trigonometric functions, $\sin ^{-1}\left(\frac{\sqrt{3}}{2} x+\frac{1}{2} \sqrt{1-x^2}\right),-\frac{1}{2}< x<\frac{1}{\sqrt{2}}$, is equal to
Let [x] denote the greatest integer less than or equal to x. Then the domain of $ f(x) = \sec^{-1}(2[x] + 1) $ is:
$(-\infty, \infty)$
$(-\infty, \infty)- \{0\}$
$(-\infty, -1] \cup [0, \infty)$
$(-\infty, -1] \cup [1, \infty)$
$\cos \left(\sin ^{-1} \frac{3}{5}+\sin ^{-1} \frac{5}{13}+\sin ^{-1} \frac{33}{65}\right)$ is equal to:
If $\alpha>\beta>\gamma>0$, then the expression $\cot ^{-1}\left\{\beta+\frac{\left(1+\beta^2\right)}{(\alpha-\beta)}\right\}+\cot ^{-1}\left\{\gamma+\frac{\left(1+\gamma^2\right)}{(\beta-\gamma)}\right\}+\cot ^{-1}\left\{\alpha+\frac{\left(1+\alpha^2\right)}{(\gamma-\alpha)}\right\}$ is equal to :
If $\frac{\pi}{2} \leq x \leq \frac{3 \pi}{4}$, then $\cos ^{-1}\left(\frac{12}{13} \cos x+\frac{5}{13} \sin x\right)$ is equal to
Using the principal values of the inverse trigonometric functions, the sum of the maximum and the minimum values of $16\left(\left(\sec ^{-1} x\right)^2+\left(\operatorname{cosec}^{-1} x\right)^2\right)$ is :
$ \text { If } y=\cos \left(\frac{\pi}{3}+\cos ^{-1} \frac{x}{2}\right) \text {, then }(x-y)^2+3 y^2 \text { is equal to } $
Explanation:
$\begin{aligned} & y=\cos \left(\frac{\pi}{3}+\cos ^{-1} \frac{x}{2}\right) \\ & =\cos \left(\frac{\pi}{3}\right) \cos \left(\cos ^{-1}\left(\frac{x}{2}\right)\right)-\sin \left(\frac{\pi}{3 .}\right) \sin \left(\cos ^{-1}\left(\frac{x}{2}\right)\right) \\ & =\frac{1}{2} \cdot \frac{x}{2}-\frac{\sqrt{3}}{2} \cdot \sqrt{1-\frac{x^2}{4}} \\ & \Rightarrow 4 y=x-\sqrt{3} \sqrt{4-x^2} \\ & \Rightarrow(4 y-x)^2=3\left(4-x^2\right) \\ & \Rightarrow 16 y^2+x^2-8 x y=12-3 x^2 \\ & x^2+4 y^2-2 x y=3 \\ & (x-y)^2+3 y^2=3 \end{aligned}$
Let S = $ \left\{ x : \cos^{-1} x = \pi + \sin^{-1} x + \sin^{-1} [2x + 1] \right\} $. Then $ \sum\limits_{x \in S} (2x - 1)^2 $ is equal to _______.
Explanation:
$\begin{aligned} & \cos ^{-1} x=\pi+\sin ^{-1} x+\sin ^{-1}(2 x+1) \\ & 2 \cos ^{-1} x-\sin ^{-1}(2 x+1)=\frac{3 \pi}{2} \\ & 2 \alpha-\beta=\frac{3 \pi}{2} \text { where } \cos ^{-1} x=\alpha, \sin ^{-1}(2 x+1)=\beta \\ & 2 \alpha=\frac{3 \pi}{2}+\beta \\ & \cos 2 \alpha=\sin \beta \\ & \begin{array}{l} 2 \cos ^2 \alpha-1=\sin \beta \\ 2 x^2-1=2 x+1 \\ x^2-x-1=0 \\ \Rightarrow x=\frac{1-\sqrt{5}}{2},\left\{x=\frac{1+\sqrt{5}}{2} \text { rejected }\right\} \\ \therefore 4 x^2-4 x=4 \\ (2 x-1)^2=5 \end{array} \end{aligned}$
If for some $\alpha, \beta ; \alpha \leq \beta, \alpha+\beta=8$ and $\sec ^2\left(\tan ^{-1} \alpha\right)+\operatorname{cosec}^2\left(\cot ^{-1} \beta\right)=36$, then $\alpha^2+\beta$ is __________
Explanation:
$\begin{aligned} & \operatorname{If}\left(\tan \left(\tan ^{-1}(\alpha)\right)+1\left(\cot \left(\cot ^{-1} \beta\right)\right)^2=36\right. \\ & \alpha^2+\beta^2=34 \\ & \alpha \beta=15 \\ & \alpha=3, \beta=5 \\ & \therefore \alpha^2+\beta=9+5=14 \end{aligned}$
Given that the inverse trigonometric function assumes principal values only. Let $x, y$ be any two real numbers in $[-1,1]$ such that $\cos ^{-1} x-\sin ^{-1} y=\alpha, \frac{-\pi}{2} \leq \alpha \leq \pi$. Then, the minimum value of $x^2+y^2+2 x y \sin \alpha$ is
If the domain of the function $\sin ^{-1}\left(\frac{3 x-22}{2 x-19}\right)+\log _{\mathrm{e}}\left(\frac{3 x^2-8 x+5}{x^2-3 x-10}\right)$ is $(\alpha, \beta]$, then $3 \alpha+10 \beta$ is equal to:
If $a=\sin ^{-1}(\sin (5))$ and $b=\cos ^{-1}(\cos (5))$, then $a^2+b^2$ is equal to
For $\alpha, \beta, \gamma \neq 0$, if $\sin ^{-1} \alpha+\sin ^{-1} \beta+\sin ^{-1} \gamma=\pi$ and $(\alpha+\beta+\gamma)(\alpha-\gamma+\beta)=3 \alpha \beta$, then $\gamma$ equals
Let $x=\frac{m}{n}$ ($m, n$ are co-prime natural numbers) be a solution of the equation $\cos \left(2 \sin ^{-1} x\right)=\frac{1}{9}$ and let $\alpha, \beta(\alpha >\beta)$ be the roots of the equation $m x^2-n x-m+ n=0$. Then the point $(\alpha, \beta)$ lies on the line
Considering only the principal values of inverse trigonometric functions, the number of positive real values of $x$ satisfying $\tan ^{-1}(x)+\tan ^{-1}(2 x)=\frac{\pi}{4}$ is :
Let the inverse trigonometric functions take principal values. The number of real solutions of the equation $2 \sin ^{-1} x+3 \cos ^{-1} x=\frac{2 \pi}{5}$, is __________.
Explanation:
$\begin{aligned} & 2 \sin ^{-1} x+3 \cos ^{-1} x=\frac{2 \pi}{5} \\ & \frac{\pi}{2}+\cos ^{-1} x=\frac{2 \pi}{5} \\ & \cos ^{-1} x=\frac{2 \pi}{5}-\frac{\pi}{2} \\ & \cos ^{-1} x=\frac{-\pi}{10} \end{aligned}$
Which is not possible as $\cos ^{-1} x \in[0, \pi]$
$\therefore \quad$ No solution
For $n \in \mathrm{N}$, if $\cot ^{-1} 3+\cot ^{-1} 4+\cot ^{-1} 5+\cot ^{-1} n=\frac{\pi}{4}$, then $n$ is equal to ________.
Explanation:
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 $
$f(x)=\log _{e}\left(4 x^{2}+11 x+6\right)+\sin ^{-1}(4 x+3)+\cos ^{-1}\left(\frac{10 x+6}{3}\right)$ is $(\alpha, \beta]$, then
$36|\alpha+\beta|$ is equal to :
Let $S = \left\{ {x \in R:0 < x < 1\,\mathrm{and}\,2{{\tan }^{ - 1}}\left( {{{1 - x} \over {1 + x}}} \right) = {{\cos }^{ - 1}}\left( {{{1 - {x^2}} \over {1 + {x^2}}}} \right)} \right\}$.
If $\mathrm{n(S)}$ denotes the number of elements in $\mathrm{S}$ then :
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 :
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 :