Question
Solve the following equation $\cos(\tan^{-1}\text{x})=\sin\Big(\cot^{-1}\frac{3}{4}\Big).$
✓
Answer
We have, $\cos(\tan^{-1}\text{x})=\sin\Big(\cot^{-1}\frac{3}{4}\Big)$L.H.S. $=\cos(\tan^{-1}\text{x})$
$=\cos\Big(\cos^{-1}\frac{1}{\sqrt{\text{x}^2+1}}\Big)$
$=\frac{1}{\sqrt{\text{x}^2+1}}$
$(\because\ \cos(\cos^{-1}\text{x})=\text{x},\ \text{x}\in[-1,1])$
R.H.S. $=\sin\Big(\cot^{-1}\frac{3}{4}\Big)$$=\sin\Big(\sin^{-1}\frac{4}{5}\Big)$
$=\frac{4}{5}$
$(\because\ \sin(\sin^{-1}\text{x})=\text{x},\ \text{x}\in[-1,1])$
$\therefore$ From given equation we get $\frac{1}{\sqrt{\text{x}^2+1}}=\frac{4}{5}$
$\Rightarrow\ 16(\text{x}^2+1)=25$
$\Rightarrow\ 16\text{x}^2=9$
$\Rightarrow\ \text{x}^2=\frac{9}{16}$
$\therefore\ \text{x}=\pm\frac{3}{4}$
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