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# 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