If $\tan ^{-1} 3+\tan ^{-1} x=\tan ^{-1} 8$, then $x=$

[question] Question. If $\tan ^{-1} 3+\tan ^{-1} x=\tan ^{-1} 8$, then $x=$ (a) 5 (b) 1/5 (c) 5/14 (d) 14/5 [/question] [solution] Solution: (b) $\frac{1}{5}$ We know that $\tan ^{-1} x+\tan ^{-1} y=\tan ^{-1} \frac{x+y}{1-x y}$. Now, $\tan ^{-1} 3+\tan ^{-1} x=\tan ^{-1} 8$ $\Rightarrow \tan ^{-1}\left(\frac{3+x}{1-3 x}\right)=\tan ^{-1} 8$ $\Rightarrow \frac{3+x}{1-3 x}=8$ $\Rightarrow 3+x=8-24 x$ $\Rightarrow 3-8=-24 x-x$ $\Rightarrow-5=-25 x$ $\Rightarrow x=\frac{5}{25}=\frac{1}{5}$ [/soluti...

If $\cos ^{-1} \frac{x}{2}+\cos ^{-1} \frac{y}{3}=\theta$, then $9 x^{2}-12 x y \cos \theta+4 y^{2}$ is equal to

[question] Question. If $\cos ^{-1} \frac{x}{2}+\cos ^{-1} \frac{y}{3}=\theta$, then $9 x^{2}-12 x y \cos \theta+4 y^{2}$ is equal to (a) 36 (b) $-36 \sin ^{2} \theta$ (c) $36 \sin ^{2} \theta$ (d) $36 \cos ^{2} \theta$ [/question] [solution] Solution: (c) $36 \sin ^{2} \theta$ We know $\cos ^{-1} x+\cos ^{-1} y=\cos ^{-1}\left[x y-\sqrt{1-x^{2}} \sqrt{1-y^{2}}\right]$ $\Rightarrow \cos ^{-1}\left[\frac{x}{2} \frac{y}{3}-\sqrt{1-\frac{x^{2}}{4}} \sqrt{1-\frac{y^{2}} {3}}\right]=\theta$ $\Rightar...

$\tan ^{-1} \frac{1}{11}+\tan ^{-1} \frac{2}{11}$ is equal to

[question] Question. $\tan ^{-1} \frac{1}{11}+\tan ^{-1} \frac{2}{11}$ is equal to (a) 0 (b) $1 / 2$ (c) $-1$ (d) none of these [/question] [solution] Solution: (d) none of these We know that $\tan ^{-1} x+\tan ^{-1} y=\tan ^{-1}\left(\frac{x+y}{1-x y}\right)$. Now, $\tan ^{-1} \frac{1}{11}+\tan ^{-1} \frac{2}{11}=\tan ^{-1}\left(\frac{\frac{1}{11}+\frac{2}{11}}{1-\frac{1} {11} \frac{2}{11}}\right)$ $=\tan ^{-1}\left(\frac{\frac{3}{11}}{\frac{121-2}{121}}\right)$ $=\tan ^{-1}\left(\frac{\frac{3}...

Let $f(x)=e^{\cos ^{-1}\{\sin (x+\pi / 3)\}}$. Then, $f(8 \pi / 9)=$

[question] Question. Let $f(x)=e^{\cos ^{-1}\{\sin (x+\pi / 3)\}}$. Then, $f(8 \pi / 9)=$ (a) $e^{5 \pi / 18}$ (b) $e^{13 \pi / 18}$ (c) $e^{-2 \pi / 18}$ (d) none of these [question] [solution] Solution: (b) $e^{13 \pi / 18}$ Given: $f(x)=e^{\cos ^{-1}\{\sin (x+\pi / 3)\}}$ Then, $f(8 \pi / 9)=e^{\cos ^{-1}\{\sin (8 \pi / 9+\pi / 3)\}}$ $=e^{\cos ^{-1}\{\sin (11 \pi / 9)\}}$ $=e^{\cos ^{-1}\left\{\cos \left(\pi / 2+{ }^{13 \pi} / 18\right)\right\}} \quad[\because \cos (\pi / 2+\theta)=\sin \the...

If $\alpha=\tan ^{-1}\left(\frac{\sqrt{3} x}{2 y-x}\right), \beta=\tan ^{-1}\left(\frac{2 x-y}{\sqrt{3} y}\right)$, then $\alpha-\beta=$

[question] Question. If $\alpha=\tan ^{-1}\left(\frac{\sqrt{3} x}{2 y-x}\right), \beta=\tan ^{-1}\left(\frac{2 x-y}{\sqrt{3} y}\right)$, then $\alpha-\beta=$ (a) $\frac{\pi}{6}$ (b) $\frac{\pi}{3}$ (C) $\frac{\pi}{2}$ (d) $-\frac{\pi}{3}$ [/question] [solution] Solution: (a) $\frac{\pi}{6}$ We have $\alpha=\tan ^{-1}\left(\frac{\sqrt{3} x}{2 y-x}\right), \beta=\tan ^{-1}\left(\frac{2 x-y}{\sqrt{3} y}\right)$ Now, $\alpha-\beta=\tan ^{-1}\left(\frac{\sqrt{3} x}{2 y-x}\right)-\tan ^{-1} \frac{2 x-...

If $\cos ^{-1} \frac{x}{3}+\cos ^{-1} \frac{y}{2}=\frac{\theta}{2}$, then $4 x^{2}-12 x y \cos \frac{\theta}{2}+9 y^{2}=$

[question] Question. If $\cos ^{-1} \frac{x}{3}+\cos ^{-1} \frac{y}{2}=\frac{\theta}{2}$, then $4 x^{2}-12 x y \cos \frac{\theta}{2}+9 y^{2}=$ (a) 36 (b) $36-36 \cos \theta$ (c) $18-18 \cos \theta$ (d) $18+18 \cos \theta$ [/question] [solution] Solution: (c) 18 − 18 cosθ We know $\cos ^{-1} x+\cos ^{-1} y=\cos ^{-1}\left(x y-\sqrt{1-x^{2}} \sqrt{1-y^{2}}\right)$ $\therefore \cos ^{-1} \frac{x}{3}+\cos ^{-1} \frac{y}{2}=\frac{\theta}{2}$ $\Rightarrow \cos ^{-1}\left(\frac{x}{3} \frac{y}{2}-\sqrt{...

If $u=\cot ^{-1} \sqrt{\tan \theta}-\tan ^{-1} \sqrt{\tan \theta}$ then, $\tan \left(\frac{\pi}{4}-\frac{u}{2}\right)=$

[question] Question. If $u=\cot ^{-1} \sqrt{\tan \theta}-\tan ^{-1} \sqrt{\tan \theta}$ then, $\tan \left(\frac{\pi}{4}-\frac{u}{2}\right)=$ (a) $\sqrt{\tan \theta}$ (b) $\sqrt{\cot \theta}$ (c) $\tan \theta$ (d) $\cot \theta$ [/question] [solution] Solution: (a) $\sqrt{\tan \theta}$ Let $y=\sqrt{\tan \theta}$ Then, $u=\cot ^{-1} \sqrt{\tan \theta}-\tan ^{-1} \sqrt{\tan \theta}$ $\Rightarrow u=\cot ^{-1} y-\tan ^{-1} y$ $\Rightarrow u=\frac{\pi}{2}-2 \tan ^{-1} y \quad\left[\because \tan ^{-1} x...

If $x<0, y<0$ such that $x y=1$, then $\tan ^{-1} x+\tan ^{-1} y$ equals

[question] Question. If $x<0, y<0$ such that $x y=1$, then $\tan ^{-1} x+\tan ^{-1} y$ equals (a) $\frac{\pi}{2}$ (b) $-\frac{\pi}{2}$ (c) $-\pi$ (d) none of these [/question] [solution] Solution: (b) $-\frac{\pi}{2}$ We know that $\tan ^{-1} x+\tan ^{-1} y=\tan ^{-1}\left(\frac{x+y}{1-x y}\right)$ $x<0, y<0$ such that $x y=1$ Let $x=-a$ and $y=-b$, where $a$ and $b$ both are positive. $\therefore \tan ^{-1} x+\tan ^{-1} y=\tan ^{-1}\left(\frac{x+y}{1-x y}\right)$ $=\tan ^{-1}\left(\frac{-a-a}{1...

If $\cos ^{-1} \frac{x}{a}+\cos ^{-1} \frac{y}{b}=\alpha$, then $\frac{x^{2}}{a^{2}}-\frac{2 x y}{a b} \cos \alpha+\frac{y^{2}}{b^{2}}=$

[question] Question. If $\cos ^{-1} \frac{x}{a}+\cos ^{-1} \frac{y}{b}=\alpha$, then $\frac{x^{2}}{a^{2}}-\frac{2 x y}{a b} \cos \alpha+\frac{y^{2}}{b^{2}}=$ (a) $\sin ^{2} \alpha$ (b) $\cos ^{2} \alpha$ (c) $\tan ^{2} \alpha$ (d) $\cot ^{2} a$ [/question] [solution] Solution: (a) $\sin ^{2} \alpha$ We know that $\cos ^{-1} x+\cos ^{-1} y=\cos ^{-1}\left(x y-\sqrt{1-x^{2}} \sqrt{1-y^{2}}\right)$. $\therefore \cos ^{-1} \frac{x}{a}+\cos ^{-1} \frac{y}{b}=\alpha$ $\Rightarrow \cos ^{-1}\left(\frac...

$2 \tan ^{-1}\left\{\operatorname{cosec}\left(\tan ^{-1} x\right)-\tan \left(\cot ^{-1} x\right)\right\}$ is equal to

[question] Question. $2 \tan ^{-1}\left\{\operatorname{cosec}\left(\tan ^{-1} x\right)-\tan \left(\cot ^{-1} x\right)\right\}$ is equal to (a) $\cot ^{-1} x$ (b) $\cot ^{-1} \frac{1}{x}$ (c) $\tan ^{-1} x$ (d) none of these [/question] [solution] Solution: (c) $\tan ^{-1} x$ Let $\tan ^{-1} x=y$ So, $x=\tan y$ $\therefore 2 \tan ^{-1}\left\{\operatorname{cosec}\left(\tan ^{-1} x\right)-\tan \left(\cot ^{-1} x\right)\right\}=2 \tan ^{-1}\left\{\operatorname{cosec}\left(\tan ^{-1} x\right)-\tan \l...

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