# Solve the following equation $\cos\big(\tan^{-1}x\big)=\sin\bigg(\cot^{-1}{\frac{3}{4}}\bigg)$

$\begin{array}{1 1} \frac{4}{3} \\ \pm \frac{3}{4} \\ \frac{4}{5} \\ \frac{3}{5} \end{array}$

Toolbox:
• $$tan^{-1}x=cos^{-1}\large\frac{1}{\sqrt{1+x^2}}$$
• $$cot^{-1}x=sin^{-1}\large\frac{1}{\sqrt{1+x^2}}$$
By taking $$x=\large\frac{3}{4},\:\large\frac{1}{\sqrt{1+x^2}}=\large\frac{1}{\sqrt{1+\large\frac{9}{16}}}=\large\frac{4}{5}$$
$$\Rightarrow\:cot^{-1}\large\frac{3}{4}=sin^{-1}\large\frac{4}{5}$$

By using the formula of $$tan^{-1}x$$ in L.H.S.
Equation becomes $$cos \bigg[ cos^{-1}\large\frac{1}{\sqrt{1+x^2}} \bigg] = sin \bigg[ sin^{-1}\large\frac{1}{\sqrt{1+\large\frac{9}{16}}} \bigg]$$
$$\Rightarrow\:cos \bigg( cos^{-1}\large\frac{1}{\sqrt{1+x^2}} \bigg) = sin \big( sin^{-1}\large\frac{4}{5}\big)$$
$$\Rightarrow \large\frac{1}{\sqrt{1+x^2}} = \large\frac{4}{5}$$

$$\Rightarrow\:1+x^2=\large\frac{25}{16}$$
$$\Rightarrow\:x^2=\large\frac{25}{16}-1=\large\frac{9}{16}$$
$$\Rightarrow\: x = \pm \large\frac{3}{4}$$

edited Mar 15, 2013