Question

The value of ${\tan }^2{27}^{\circ }+2\tan {27}^{\circ }\tan {36}^{\circ }$ is equal to

Answer 1

${\tan }^2{27}^{\circ }+2\tan {27}^{\circ }\tan {36}^{\circ }$

$=(1+{\tan }^2{27}^{\circ })+\frac{2\tan {27}^{\circ }\tan {36}^{\circ }}{1+{\tan }^2{27}^{\circ }}\times (1+{\tan }^2{27}^{\circ })-1$

$=(1+{\tan }^2{27}^{\circ })(1+\frac{2\tan {27}^{\circ }}{1+{\tan }^2{27}^{\circ }}.\tan {36}^{\circ })-1$

$={\sec }^2{27}^{\circ }(1+\sin {54}^{\circ }.\tan {36}^{\circ })-1$

$={\sec }^2{27}^{\circ }(1+\sin {36}^{\circ })-1$

$={\sec }^2{27}^{\circ }(1+\cos {54}^{\circ })-1$

$=2-1=1$