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.