Prove that: ( 4 +x)+ ( 4 -x)=2 2x - Vidyadip

Question

Prove that:
$\tan(\frac{\pi}{4}+\text{x})+\tan(\frac{\pi}{4}-\text{x})=2\sec2\text{x}$

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Answer

$\text{LHS}=\tan\Big(\frac{\pi}{4}+\text{x}\Big)+\tan\frac{\pi}{4}-\text{x}\Big)$
$=\frac{\tan\frac{\pi}{4}+\tan\text{x}}{1-\tan\frac{\pi}{4}\tan\text{x}}+\frac{\tan\frac{\pi}{4}-\tan\text{x}}{1+\tan\frac{\pi}{4}\tan\text{x}}$
$=\frac{1+\tan\text{x}}{1-\tan\text{x}}+\frac{1-\tan\text{x}}{1+\tan\text{x}}$ $\Big[\because\tan\frac{\pi}{4}=1\Big]$
$=\frac{(1+\tan^2\text{x}+2\text{x})+(1+\tan^2\text{x}-1\tan\text{x})}{(1-\tan\text{x})(1+\tan\text{x})}$
$=\frac{1(1+\tan^2\text{x})}{1-\tan^2\text{x}}$
$=\frac{1\sec^2\text{x}}{1-\frac{\sin^2\text{x}}{\cos^2\text{x}}}$ $[\because\sec^2\text{x}=1+\tan^2\text{x}]$
$=\frac{2\sec^2\text{x}.\cos^2\text{x}}{\cos^2\text{x}-\sin62\text{x}}$ $[\because\sec-\frac{1}{\cos\text{x}}]$
$=\frac{2}{\cos2\text{x}}$
$=2\sec2\text{x}=\text{RHS}$

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