Question

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

Answer 1

$\frac{\tan {155}^{\circ }-\tan {115}^{\circ }}{1+\tan {155}^{\circ }\tan {115}^{\circ }}=\tan (155-115{)}^{\circ }$

$=\tan {40}^{\circ }$

$=\tan (90-50{)}^{\circ }$

$=\cot {50}^{\circ }$

$=\frac{1}{\tan {50}^{\circ }}$

$=\frac{1}{\tan 2(25{)}^{\circ }}$

$=\frac{1}{\frac{2\tan {25}^{\circ }}{1-{\tan }^2{25}^{\circ }}}$ ( $\tan 2A=\frac{2\tan A}{1-{\tan }^{2}A}$ )

$=\frac{1}{\frac{2a}{1-a^2}}$

$=\frac{1-a^2}{2a}$