Question

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

Answer 1

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

$Letta{n}^{-1}\frac{1}{4}=x\Rightarrow tanx=\frac{1}{4}\Rightarrow ta{n}^{2}x=\frac{1}{16}\Rightarrow se{c}^{2}x-1=\frac{1}{16}\Rightarrow se{c}^{2}x=1+\frac{1}{16}=\frac{17}{16}\Rightarrow \frac{1}{co{s}^{2}x}=\frac{17}{16}\Rightarrow co{s}^{2}x=\frac{16}{17}\Rightarrow cosx=\frac{4}{\sqrt{17}}$

$\Rightarrow si{n}^{2}x=1-co{s}^{2}x=1-\frac{16}{17}=\frac{1}{17}$
$\Rightarrow sinx=\frac{1}{\sqrt{17}}Again,letta{n}^{-1}\frac{2}{9}=y\Rightarrow tany=\frac{2}{9}\Rightarrow ta{n}^{2}y=\frac{4}{81}$

$\Rightarrow se{c}^{2}y-1=\frac{4}{81}\Rightarrow se{c}^{2}y=\frac{4}{81}+1=\frac{85}{81}\Rightarrow co{s}^{2}y-=\frac{81}{85}\Rightarrow cosy=\frac{9}{\sqrt{85}}\Rightarrow si{n}^{2}y=1-co{s}^{2}y=1-\frac{81}{85}=\frac{4}{85}\Rightarrow siny=\frac{2}{\sqrt{85}}$

We know that, $sin(x+y)=sinx.cosy+cosx.siny=\frac{1}{\sqrt{17}}.\frac{9}{\sqrt{85}}+\frac{4}{\sqrt{17}}.\frac{2}{\sqrt{85}}=\frac{17}{\sqrt{17}.\sqrt{85}}=\frac{\sqrt{17}}{\sqrt{17}.\sqrt{5}}=\frac{1}{\sqrt{5}}\Rightarrow (x+y)=si{n}^{-1}\frac{1}{\sqrt{5}}\Rightarrow ta{n}^{-1}\frac{1}{4}+ta{n}^{-1}\frac{2}{9}=si{n}^{-1}\frac{1}{\sqrt{5}}Henceproved.$