Question

If $\tan {25}^{\circ }=x,$ prove that $\frac{\tan {155}^{\circ }-\tan {115}^{\circ }}{1+\tan {155}^{\circ }\tan {115}^{\circ }}=\frac{1-x^2}{2x}$.

Answer 1

$\tan {25}^{\circ }=x$

$L.H.S=\frac{\tan {155}^{\circ }-\tan {115}^{\circ }}{1+\tan {155}^{\circ }\tan {115}^{\circ }}$

$=\frac{\tan \left(180-25\right)-\tan \left(90+25\right)}{1+\tan \left(180-25\right)\tan \left(90+25\right)}$

$=\frac{-\tan 25+\cot 25}{1+\tan 25\cot 25}$

$\frac{-x+1/x}{1+x/x}=\frac{-x^2+1}{2}=\frac{1-x^2}{2}$