यदि $\tan ^{-1}\left(\frac{1}{4}\right)+\tan ^{-1}\left(\frac{2}{9}\right)=\frac{1}{2} \cos ^{-1} x$ है,तो $x$ का मान ज्ञात कीजिए।

  • A
    $\frac{1}{5}$
  • B
    $\frac{2}{5}$
  • C
    $\frac{3}{5}$
  • D
    $\frac{4}{5}$

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Similar Questions

यदि $\pi \leq x \leq 2 \pi$ है,तो $\cos^{-1}(\cos x)$ किसके बराबर है :-

$\cot ^{-1}\left(\frac{\sqrt{1+\tan ^2(2)}-1}{\tan (2)}\right)-\cot ^{-1}\left(\frac{\sqrt{1+\tan ^2\left(\frac{1}{2}\right)}+1}{\tan \left(\frac{1}{2}\right)}\right)$ का मान ज्ञात कीजिए।

कथन-$1$: ${\cot ^{ - 1}}\left[ {\frac{{\log (e/{x^2})}}{{\log (ex^2)}}} \right] + {\cot ^{ - 1}}\left[ {\frac{{\log (ex^2)}}{{\log (e/{x^2})}}} \right] = \frac{\pi}{2}$
कथन-$2$: ${\tan ^{ - 1}}\left[ {\frac{{1 + \log {x^2}}}{{1 - \log {x^2}}}} \right] = {\tan ^{ - 1}}1 + {\tan ^{ - 1}}(\log {x^2})$

यदि त्रिभुज $ABC$ में,$A = \tan^{-1} 2$ और $B = \tan^{-1} 3$ है,तो कोण $C$ का मान क्या होगा?

सिद्ध कीजिए कि $\tan ^{-1} \sqrt{x} = \frac{1}{2} \cos ^{-1} \left( \frac{1-x}{1+x} \right)$,जहाँ $x \in [0, 1]$ है।

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