Question

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

Answer 1

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

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

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

${\sec }^2\theta \cdot \frac{1}{{\sec }^2\theta }=k$

$1=k$

$\therefore k=1$.