Question

Solve the following equation $cos\left(ta{n}^{-1}x\right)=sin\left(co{t}^{-1}\frac{3}{4}\right)$

Answer 1

$\cos \left(ta{n}^{-1}x\right)=\sin \left({\cot }^{-1}\frac{3}{4}\right)\therefore let{\theta }_1={\tan }^{-1}x\&{\theta }_2={\cot }^{-1}\frac{3}{4}\therefore x=\tan \theta ,\&\frac{3}{4}=\cot {\theta }_2\Rightarrow \frac{\sin {\theta }_1}{\cos {\theta }_1}=x\&\frac{\cos {\theta }_2}{\sin {\theta }_2}=\frac{3}{4}\Rightarrow \frac{{\sin }^2{\theta }_1}{{\cos }^2{\theta }_1}=x^2\&\frac{{\cos }^2{\theta }_2}{{\sin }^2{\theta }_2}=\frac{9}{16}\Rightarrow \frac{1-{\cos }^2{\theta }_1}{{\cos }^2{\theta }_1}=x^2\&\frac{1-{\sin }^2{\theta }_2}{{\sin }^2{\theta }_2}=\frac{9}{16}\Rightarrow \cos {\theta }_1=\frac{1}{\sqrt{1+x^2}}\&\sin {\theta }_2=\frac{4}{5}\therefore {\theta }_1={\cos }^{-1}\frac{1}{\sqrt{1+x^2}}\&{\theta }_2={\sin }^{-1}\frac{4}{5}\therefore \cos \left({\cos }^{-1}\frac{1}{\sqrt{1+x^2}}\right)=\sin \left({\sin }^{-1}\frac{4}{5}\right)\Rightarrow \frac{1}{\sqrt{1+x^2}}=\frac{4}{5}(\because \cos {\cos }^{-1}\theta =\theta \&\sin {\sin }^{-1}\theta =\theta )\therefore x^2+1=\frac{25}{16}\therefore x=\frac{3}{4}$