Question

if sec²theta (1+sin) (1-sin theta)= k, then find the value of k

Answer 1

$$\begin{gathered} {\sec }^2\mathrm{\theta }\left(1+\sin \mathrm{\theta }\right)\left(1-\sin \mathrm{\theta }\right)=\mathrm{k} \\ \frac{\left(1+\sin \mathrm{\theta }\right)\left(1-\sin \mathrm{\theta }\right)}{{\cos }^2\mathrm{\theta }}=\mathrm{k} \\ \frac{\left(1+\sin \mathrm{\theta }\right)\left(1-\sin \mathrm{\theta }\right)}{\left(1-{\sin }^2\mathrm{\theta }\right)}=\mathrm{k}\left(\mathrm{Since}{\cos }^2\mathrm{\theta }=1-{\sin }^2\mathrm{\theta }\right) \\ \frac{\left(1+\sin \mathrm{\theta }\right)\left(1-\sin \mathrm{\theta }\right)}{\left(1+\sin \mathrm{\theta }\right)\left(1-\sin \mathrm{\theta }\right)}=\mathrm{k} \\ \mathrm{k}=1\mathrm{ANS}... \\ \mathrm{Value}\mathrm{of}\mathrm{k}\mathrm{is}1. \end{gathered}$$