in Quantitative Aptitude recategorized by
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$ \ Sin^{-1}\left [ \frac{3}{5} \right ] + \tan^{-1}\left [ \frac{1}{7} \right ]=$

  1. $\frac{\pi }{4}$
  2. $\frac{\pi }{2}$
  3. $ \ Cos^ {-1} \frac{4}{5} $
  4. $\pi$
in Quantitative Aptitude recategorized by
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$\textrm{let us assume $sin^{-1}(\frac{3}{5})=\theta$}$

$\implies$ $\sin\theta=\frac{3}{5}$

$\because \sin\theta=\frac{perpendicular}{base}$

$\textrm{by using pythagoras theorem we can find the base of right angle triangle }$

$\implies$ $H^2=B^2+P^2$

$\textrm{base =4}$

$\therefore$ $\tan\theta=\frac{perpendicular}{base}$

$\implies$ $\tan\theta=\frac{3}{4}$

$\implies$ $\theta=\tan^{-1}\frac{3}{4}$

$\therefore$ $tan^{-1}\frac{3}{4}+tan^{-1}\frac{1}{7}$

$\because \tan^{-1}(A+B)=tan^{-1}(\frac{A+B}{1-AB})$

$\implies$ $\tan^{-1}(\frac{\frac{3}{4}+\frac{1}{7}}{1-\frac{3}{4}*\frac{1}{7}})$

$\implies$ $tan^{-1}(1)$

$\because tan^{-1}(1)=\frac{\pi}{4}$

$\textrm{option A is correct.}$
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