Question

tan(π/4+x)-tan(π/4-x)=2tan 2x

Answer 1

$\textbf{To prove:}$

$tan(\frac{\pi}{4}+x)-tan(\frac{\pi}{4}-x)=2\,tan2x$

$\textbf{Solution:}$

$\text{Consider,}$

$tan(\frac{\pi}{4}+x)-tan(\frac{\pi}{4}-x)$

$\text{Using the following identities,}$

$\boxed{\bf\,tan(A+B)=\frac{tanA+tanB}{1-tanA\,tanB}}$

$\boxed{\bf\,tan(A-B)=\frac{tanA-tanB}{1+tanA\,tanB}}$

$=(\dfrac{tan\frac{\pi}{4}+tanx}{1-tan\frac{\pi}{4}\,tanx})-(\dfrac{tan\frac{\pi}{4}-tanx}{1+tan\frac{\pi}{4}\,tanx}})$

$=(\dfrac{1+tanx}{1-tanx})-(\dfrac{1-tanx}{1+tanx}})$

$=\dfrac{(1+tanx)(1+tanx)-(1+tanx)(1-tanx)}{(1-tanx)(1+tanx)}$

$=\dfrac{(1+tanx)^2-(1-tanx)^2}{1-tan^2x}$

$=\dfrac{1+tan^2x+2\,tanx-1-tan^2x+2\,tanx}{1-tan^2x}$

$=\dfrac{4\,tanx}{1-tan^2x}$

$=2(\dfrac{2\,tanx}{1-tan^2x})$

$\text{Using the identity}$

$\boxed{\bf\,tan2A=\frac{2\,tanA}{1-tan^2A}}$

$=2\,tan2x$

$\therefore\bf\,tan(\frac{\pi}{4}+x)-tan(\frac{\pi}{4}-x)=2\,tan2x$

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Answer 2

Please see the image for your soln

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