Question

Show that $\tan {15}^o+\tan {75}^o=\frac{{\sec }^2{15}^o}{\sqrt{{\sec }^2{15}^o-1}}$.

Answer 1

Now,

$\tan {15}^o+\tan {75}^o$

$=\tan {15}^o+\cot {15}^o$ [ Since $\tan {75}^o=\cot {15}^o$]

$=\tan {15}^o+\frac{1}{\tan {15}^o}$

$=\frac{1+{\tan }^2{15}^o}{\tan {15}^o}$

$=\frac{{\sec }^2{15}^o}{\sqrt{{\sec }^2{15}^o-1}}$ [ As ${\sec }^2{15}^o=1+{\tan }^2{15}^o$]