If log_{pi}x 0,then the value of log_{pi}le | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) Given that $\log_{\pi}x > 0$. Since the base $\pi > 1$,this implies $x > 1$.We know the identity for $\sin^{-1}\left(\frac{2x}{1+x^2}\right)$:For

Vedclass · Accepted Answer

$1$

Vedclass · Answer

(C) Given that $\log_{\pi}x > 0$. Since the base $\pi > 1$,this implies $x > 1$.We know the identity for $\sin^{-1}\left(\frac{2x}{1+x^2}ight)$:For $x > 1$,$\sin^{-1}\left(\frac{2x}{1+x^2}ight) = \pi - 2	an^{-1}x$.Substituting this into the expression:$\log_{\pi}\left( \sin^{-1}\frac{2x}{1+x^2} + 2	an^{-1}x ight) = \log_{\pi}\left( (\pi - 2	an^{-1}x) + 2	an^{-1}x ight)$.Simplifying the expression inside the logarithm:$\log_{\pi}(\pi) = 1$.

If $\log_{\pi}x > 0$,then the value of $\log_{\pi}\left( \sin^{-1}\frac{2x}{1+x^2} + 2\tan^{-1}x \right)$ is equal to

Similar Questions

$\tan \left[ {\frac{\pi }{4} + \frac{1}{2}{{\cos }^{ - 1}}\frac{a}{b}} \right] + \tan \left[ {\frac{\pi }{4} - \frac{1}{2}{{\cos }^{ - 1}}\frac{a}{b}} \right] = $

If $x=\operatorname{cosec}(\tan ^{-1}(\cos (\cot ^{-1}(\sec (\sin ^{-1} a)))))$,where $a \in [0, 1]$,then which of the following is true?

If $\theta = \sin^{-1}x + \cos^{-1}x - \tan^{-1}x$ for $x \ge 0$,then the smallest interval in which $\theta$ lies is given by

Prove that $2 \tan ^{-1} \frac{1}{2} + \tan ^{-1} \frac{1}{7} = \tan ^{-1} \frac{31}{17}$.

$(\tan ^{-1} x)^2+(\cot ^{-1} x)^2=\frac{5 \pi^2}{8} \Rightarrow x=$

Vedclass Test Series

Exam Paper Generator

Online Exam Module