Question

${\tan }^{-1}\frac{1}{4}+{\tan }^{-1}\frac{2}{9}={\sin }^{-1}\frac{1}{\sqrt{5}}$

Answer 1

R.E.F.Image

$ta{n}^{-1}\frac{1}{4}+ta{n}^{-1}\frac{2}{9}=si{n}^{-1}\frac{1}{\sqrt{5}}$

L.H.S $ta{n}^{-1}\frac{1}{4}+ta{n}^{-1}\frac{2}{9}$ $\because ta{n}^{-1}A+ta{n}^{-1}B=ta{n}^{-1}\left(\frac{A+B}{1-AB}\right)$

$\Rightarrow ta{n}^{-1}\left(\frac{\frac{1}{4}+\frac{2}{9}}{1-\frac{2}{36}}\right)$

$\Rightarrow ta{n}^{-1}\left(\frac{\frac{9+8}{36}}{\frac{34}{36}}\right)$

$\Rightarrow ta{n}^{-1}\left(\frac{17}{34}\right)\Rightarrow ta{n}^{-1}\left(\frac{1}{2}\right)=\theta$ (Let)

$tan\theta =\frac{1}{2}\frac{\left(perpendicular\right)}{base}$

$sin\theta =\frac{perpendicular}{Hypotenuse}$

$sin\theta =\frac{1}{\sqrt{5}}$

$\theta =si{n}^{-1}\frac{1}{\sqrt{5}}$ R.H.S