Question

If $k+1={\sec }^2\theta (1+\sin \theta )(1-\sin \theta ),$ then find the value of $k$.

Answer 1

Given

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

$\Rightarrow K+1={\sec }^2\theta (1-{\sin }^2\theta )$ (Using $(a-b)(a+b)=a^2-b^2$)

$\Rightarrow K+1={\sec }^2\theta .{\cos }^2\theta$ ($\because$ $1-{\sin }^2\theta ={\cos }^2\theta$)

$\Rightarrow K+1=1$

$\therefore$ $K=0$