Question

If $\frac{1}{1+\sin \mathrm{\theta }}+\frac{1}{1-\sin \mathrm{\theta }}=k{\sec }^2\mathrm{\theta }$, then the value of k is _________.

Answer 1

$$\begin{gathered} \frac{1}{1+\sin \theta }+\frac{1}{1-\sin \theta } \\ =\frac{1-\sin \theta +1+\sin \theta }{\left(1+\sin \theta \right)\left(1-\sin \theta \right)} \\ =\frac{2}{1-{\sin }^2\theta } \end{gathered}$$
$$\begin{gathered} =\frac{2}{{\cos }^2\theta }\left({\cos }^2\theta +{\sin }^2\theta =1\right) \\ =2{\sec }^2\theta \end{gathered}$$

Comparing with the given expression, we get

k = 2

If $\frac{1}{1+\sin \mathrm{\theta }}+\frac{1}{1-\sin \mathrm{\theta }}=k{\sec }^2\mathrm{\theta }$, then the value of k is ____2____.