If $x, y, z$ are in Arithmetic Progression and $\tan^{-1} x, \tan^{-1} y, \tan^{-1} z$ are also in Arithmetic Progression,where $x, z > 0$ and $xz < 1, y < 1$,then
- A$x = y = z$
- B$2x = 3y = 6z$
- C$6x = 3y = 2z$
- D$6x = 4y = 3z$
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View SolutionIf the equation $\sin^{-1} \sqrt{x} + \cos^{-1} \sqrt{x^2 - 1} + \tan^{-1} (\tan y) = a$ has at least one solution,then the number of integral values of $a$ is:
Match List $I$ with List $II$ and select the correct answer using the code given below the lists:
List $I$ List $II$ $P$. $\left(\frac{1}{y^2}\left(\frac{\cos (\tan ^{-1} y)+y \sin (\tan ^{-1} y)}{\cot (\sin ^{-1} y)+\tan (\sin ^{-1} y)}\right)^2+y^4\right)^{1 / 2}$ takes value $1$. $\frac{1}{2} \sqrt{\frac{5}{3}}$ $Q$. If $\cos x+\cos y+\cos z=0=\sin x+\sin y+\sin z$ then possible value of $\cos \frac{x-y}{2}$ is $2$. $\sqrt{2}$ $R$. If $\cos (\frac{\pi}{4}-x) \cos 2 x+\sin x \sin 2 x \sec x=\cos x \sin 2 x \sec x+\cos (\frac{\pi}{4}+x) \cos 2 x$ then possible value of $\sec x$ is $3$. $\frac{1}{2}$ $S$. If $\cot (\sin ^{-1} \sqrt{1-x^2})=\sin (\tan ^{-1}(x \sqrt{6})), x \neq 0$,then possible value of $x$ is $4$. $1$
Codes: $P \quad Q \quad R \quad S$
| List $I$ | List $II$ |
|---|---|
| $P$. $\left(\frac{1}{y^2}\left(\frac{\cos (\tan ^{-1} y)+y \sin (\tan ^{-1} y)}{\cot (\sin ^{-1} y)+\tan (\sin ^{-1} y)}\right)^2+y^4\right)^{1 / 2}$ takes value | $1$. $\frac{1}{2} \sqrt{\frac{5}{3}}$ |
| $Q$. If $\cos x+\cos y+\cos z=0=\sin x+\sin y+\sin z$ then possible value of $\cos \frac{x-y}{2}$ is | $2$. $\sqrt{2}$ |
| $R$. If $\cos (\frac{\pi}{4}-x) \cos 2 x+\sin x \sin 2 x \sec x=\cos x \sin 2 x \sec x+\cos (\frac{\pi}{4}+x) \cos 2 x$ then possible value of $\sec x$ is | $3$. $\frac{1}{2}$ |
| $S$. If $\cot (\sin ^{-1} \sqrt{1-x^2})=\sin (\tan ^{-1}(x \sqrt{6})), x \neq 0$,then possible value of $x$ is | $4$. $1$ |
Codes: $P \quad Q \quad R \quad S$
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