Question

Statement I The valuc of $\tan ^{-1} \frac{1}{4}+\tan ^{-1} \frac{2}{9}$ is equal to $\sin ^{-1} \frac{1}{\sqrt{5}}$
Statement II $\sin ^{-1} \frac{8}{17}+\sin ^{-1} \frac{3}{5}=\sin ^{-1} \frac{77}{85}$.

• A. Only I is true
• B. Only II is true
• C. Both I and II are true
• D. Neither I nor II is true

Answer 1

Answer: (C)

I. Consider, $ \tan ^{-1} \frac{1}{4}+\tan ^{-1} \frac{2}{9}$
$ =\tan ^{-1}\left(\frac{\frac{1}{4}+\frac{2}{9}}{1-\frac{1}{4} \times \frac{2}{9}}\right)=\tan ^{-1}\left(\frac{9+8}{36-2}\right)$
$=\tan ^{-1}\left(\frac{17}{34}\right)=\tan ^{-1}\left(\frac{1}{2}\right)$
Now, let $ \tan ^{-1} \frac{1}{2}=\theta \Rightarrow \tan \theta=\frac{1}{2} $
$ \therefore \sec \theta=\sqrt{1+\left(\frac{1}{2}\right)^2} \left(\because \sec \theta=\sqrt{1+\tan ^2 \theta}\right)$
$ =\frac{\sqrt{5}}{2}$
$ \cos \theta=\frac{2}{\sqrt{5}}$
$ \sin \theta=\sqrt{1-\frac{4}{5}} \left(\because \sin \theta=\sqrt{1-\cos ^2 \theta}\right) $
$ =\frac{1}{\sqrt{5}} \Rightarrow \theta=\sin ^{-1} \frac{1}{\sqrt{5}} $
$\therefore$ Statement I is true.
II. $LHS =\sin ^{-1} \frac{8}{17}+\sin ^{-1} \frac{3}{5} $
$=\sin ^{-1}\left[\frac{8}{17} \sqrt{1-\left(\frac{3}{5}\right)^2}+\frac{3}{5} \sqrt{1-\left(\frac{8}{17}\right)^2}\right] $
$ {\left[\because \sin ^{-1} x+\sin ^{-1} y=\sin ^{-1}\left(x \sqrt{1-y^2}+y \sqrt{1-x^2}\right)\right] }$
$=\sin ^{-1}\left[\frac{8}{17} \times \frac{4}{5}+\frac{3}{5} \times \frac{15}{17}\right] $
$=\sin ^{-1}\left[\frac{32+45}{85}\right]=\sin ^{-1} \frac{77}{85}=\text { RHS }$
Thus, statement II is also true.