Question

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

Answer 1

We have cos$\left(ta{n}^{-1}x\right)=sin\left(co{t}^{-1}\frac{3}{4}\right)$

$\Rightarrow cos\left(co{s}^{-1}\frac{1}{\sqrt{x^2+1}}\right)=sin\left(si{n}^{-1}\frac{4}{5}\right)(Letta{n}^{-1}x={\theta }_1\Rightarrow tan{\theta }_1=\frac{x}{1}\Rightarrow cos{\theta }_1=\frac{1}{\sqrt{x^2+1}}\Rightarrow {\theta }_1=co{s}^{-1}\frac{1}{\sqrt{x^2+1}}andco{t}^{-1}\frac{3}{4}={\theta }_{2}cot{\theta }_2=\frac{3}{4}\Rightarrow sin{\theta }_2=\frac{4}{5}\Rightarrow {\theta }_2=si{n}^{-1}\frac{4}{5}\Rightarrow \frac{1}{\sqrt{x^2+1}}=\frac{4}{5}\{\because cos(co{s}^{-1}x)=x,x\in [-1,1\left]\text{and sin}\right(si{n}^{-1}x)=x,x\in [-1,1\left]\right\}$

On squaring both sides, we get

$16(x^2+1)=25\Rightarrow 16x^2=9\Rightarrow x^2={\left(\frac{3}{4}\right)}^2\therefore x=\pm \frac{3}{4}=\frac{-3}{4},\frac{3}{4}$