Question

Prove that $\left(1+{\tan }^2\theta \right)\left(1+\sin \theta \right)\left(1-\sin \theta \right)=1$

Answer 1

L.H.S

$(1+{\tan }^2\theta )(1+\sin \theta )(1-\sin \theta )$

Since, $x^2-y^2=(x+y)(x-y)$

Therefore,

$=(1+{\tan }^2\theta )(1-{\sin }^2\theta )$

Since,

${\sec }^{2}x=1+{\tan }^{2}x$

${\cos }^{2}x=1-{\sin }^{2}x$

Therefore,

$={\sec }^2\theta \times {\cos }^2\theta$

$=\frac{1}{{\cos }^2\theta }\times {\cos }^2\theta$

$=1$

Hence, proved.