Question

Show that :$(\sec \mathrm{\theta }+\cos \mathrm{\theta })(\sec \mathrm{\theta }-\cos \mathrm{\theta })={\tan }^2\mathrm{\theta }+{\sin }^2\mathrm{\theta }$

Answer 1

To prove :

$(\sec \mathrm{\theta }+\cos \mathrm{\theta })(\sec \mathrm{\theta }-\cos \mathrm{\theta })={\tan }^2\mathrm{\theta }+{\sin }^2\mathrm{\theta }$

Proof:

$$\begin{gathered} \mathrm{LHS}=(\mathrm{sec\theta }+\mathrm{cos\theta })(\mathrm{sec\theta }-\mathrm{cos\theta }) \\ =\sec {}^2\mathrm{\theta }-{\cos }^2\mathrm{\theta }\left[\because \left(\mathrm{a}+\mathrm{b}\right)(\mathrm{a}-\mathrm{b})={\mathrm{a}}^2-{\mathrm{b}}^2\right] \\ =(1+{\tan }^2\mathrm{\theta })-(1-{\sin }^2\mathrm{\theta })\left[\because \left({\sec }^2\mathrm{\theta }-{\tan }^2\mathrm{\theta }=1\right)\&({\sin }^2\mathrm{\theta }+{\cos }^2\mathrm{\theta }=1)\right] \\ =1+{\tan }^2\mathrm{\theta }-1+{\sin }^2\mathrm{\theta } \\ ={\tan }^2\mathrm{\theta }+{\sin }^2\mathrm{\theta } \\ =\mathrm{RHS} \end{gathered}$$

Thus, $(\sec \mathrm{\theta }+\cos \mathrm{\theta })(\sec \mathrm{\theta }-\cos \mathrm{\theta })={\tan }^2\mathrm{\theta }+{\sin }^2\mathrm{\theta }$.

Hence Proved.