Inverse Trigonometric Functions · Mathematics · AP EAPCET
MCQ (Single Correct Answer)
If $\theta=\tan ^{-1}\left(\frac{1}{3}\right)+\tan ^{-1}\left(\frac{1}{7}\right)+\tan ^{-1}\left(\frac{1}{13}\right) +\tan ^{-1}\left(\frac{1}{21}\right)+\tan ^{-1}\left(\frac{1}{31}\right)$, then $\tan \theta=$
If $\tan ^{-1} x=\cot h^{-1} y=\log \sqrt{5}$, then $\tan ^{-1}(x y)=$
If $f(x)=2+\left|\sin ^{-1} x\right|$ and $A=\left\{x \in R / f^1(x)\right.$ exists $\}$, then $A=$
The equation $\cos ^{-1}(1-x)-2 \cos ^{-1} x=\frac{\pi}{2}$ has
$$ \tan \left(2 \tan ^{-1}\left(\frac{1}{3}\right)+\tan ^{-1}\left(\frac{1}{7}\right)\right)= $$
$$ \tanh ^{-1}\left(\frac{1}{3}\right)+\operatorname{coth}^{-1}(3)= $$
If $y=\sin ^{-1}\left(\frac{2 x}{1+x^2}\right)$ and $\left(\frac{d^2 y}{d x^2}\right)_{x=2}=k$, then $25 k=$
If $f(x)=\sec ^{-1}\left(\frac{1}{2 x^2-1}\right)$ and $g(x)=\tan ^{-1}\left(\frac{\sqrt{1+x^2}-1}{x}\right)$, then the derivative of $f(x)$ with respect to $g(x)$ is
The domain of the function, $f(x)=\sqrt{\log _e\left(\frac{1}{x^2-4 x+4}\right)}+\sin ^{-1}\left(x^2-2\right)$ is
If $\cot \left(\cos ^{-1} x\right)=\sec \left\{\tan ^{-1}\left(\frac{a}{\sqrt{b^2-a^2}}\right)\right\}, b>a$ then $x=$
If $\sinh ^{-1} x=\log 3$ and $\cosh ^{-1} y=\log \frac{3}{2}$, then $\tanh ^{-1}(x-y)=$
The number of solution of $\tan ^{-1} 1+\frac{1}{2} \cos ^{-1} x^2-\tan ^{-1} \left(\frac{\sqrt{1+x^2}+\sqrt{1-x^2}}{\sqrt{1+x^2}-\sqrt{1-x^2}}\right)=0$ is
$$ \tanh ^{-1}(\sin \theta)= $$
The interval in which the function $f(x)=\tan ^{-1}(\sin x+\cos x)$ is an increasing function is
The range of the real valued function $f(x)=\cos ^{-1}\left(\frac{3}{\sqrt{9 x^2-12 x+22}}\right)$ is
If the equation $2 \cot ^{-1}\left(x^2+2 x+k\right)=\pi-3 \tan ^{-1} \left(x^2+2 x+k\right)$ has two distinct real solutions, then all the values of $k$ lie in the interval
$$ \sec h^{-1}(\sin \alpha)= $$
If $y=\log \left(\sec \left(\tan ^{-1} x\right)\right)(x>0)$, then $\frac{d y}{d x}$ at $x=1$ is
If $y=\sin ^{-1} \frac{\sqrt{1+\sin x}+\sqrt{1-\sin x}}{\sqrt{1+\sin x}-\sqrt{1-\sin x}}$ and $\frac{-3 \pi}{2}
If $\frac{1}{2} \sin ^{-1}\left(\frac{3 \sin 2 \theta}{5+4 \cos 2 \theta}\right)=\tan ^{-1} x$, then $x=$
If $\operatorname{sech}^{-1} x=\log 2$ and $\operatorname{cosech}^{-1} y=-\log 3$, then $(x+y)=$
If $y=\tan ^{-1}\left(\frac{x}{1+2 x^2}\right)+\tan ^{-1}\left(\frac{x}{1+6 x^2}\right)$, then $\frac{d y}{d x}=$
The range of the real valued function $f(x)=\cos ^{-1}(-x)+\sin ^{-1}(-x)+\operatorname{cosec}^{-1}(x)$ is
The horizontal distance between a tower and a building is $10 \sqrt{3}$ units. If the angle of depression of the foot of the building from the top of the tower is $60^{\circ}$ and the angle of elevation of the top of the building from the foot of the tower is $30^{\circ}$, then the sum of the heights of the tower and the building is
If $x$ is a real number, then the number of solutions of $\tan ^{-1}(\sqrt{x(x+1)})+\sin ^{-1}\left(\sqrt{x^2+x+1}\right)=\frac{\pi}{2}$ is
If $y=\tanh ^{-1} \sqrt{\frac{1-x}{1+x}}$, then $\frac{d y}{d x}=$
$$ \tan ^{-1} \frac{\sqrt{8-2 \sqrt{15}}}{\sqrt{15}+1}+\tan ^{-1} \frac{1}{\sqrt{5}}= $$
The derivative of $\sec ^{-1}\left(\frac{1}{2 x^2-1}\right)$ with respect to $\sqrt{1-x^2}$ at $x=\frac{1}{2}$ is
If $0 < x < \frac{1}{2}$ and $\alpha=\sin ^{-1} x+\cos ^{-1}\left(\frac{x}{2}+\frac{\sqrt{3-3 x^2}}{2}\right)$, then $\tan \alpha+\cot \alpha$ is equal to
$$\tan ^{-1}(-2)-\tan ^{-1}(3)$$ is equal to
If $$x=\sin \left(2 \tan ^{-1} 2\right), y=\cos \left(2 \tan ^{-1} 3\right)$$ and $$z=\sec \left(3 \tan ^{-1} 4\right)$$, then
$$\frac{d}{d x}\left\{\sin ^2\left(\cot ^{-1} \sqrt{\frac{1+x}{1-x}}\right)\right\}$$ is equal to
If $$y=\tan ^{-1}\left\{\frac{a x-b}{b x+a}\right\}$$, then $$y^{\prime}$$ is equal to
For how many distinct values of $$x$$, the following $$\sin \left[2 \cos ^{-1} \cot \left(2 \tan ^{-1} x\right)\right]=0$$ holds?
If $$\tan ^{-1}\left[\frac{1}{1+1 \cdot 2}\right]+\tan ^{-1}\left[\frac{1}{1+2 \cdot 3}\right]+\ldots+\tan ^{-1} \left[\frac{1}{1+n(1+1)}\right]=\tan ^{-1}[x]$$, then $$x$$ is equal to
If $$y=\tan ^{-1}\left(\frac{\sqrt{1+x^2}+\sqrt{1-x^2}}{\sqrt{1+x^2}-\sqrt{1-x^2}}\right)$$, where $$x^2 \leq 1$$. Then, find $$\frac{d y}{d x}$$ is equal to
If $$\int \frac{d x}{x\left(\sqrt{\left.x^4-1\right)}\right.}=\frac{1}{k} \sec ^{-1}\left(x^k\right)$$, then the value of $$k$$ is equal to