Question

Prove the following identities (1-17)

${\left(1+\tan \alpha \tan \mathrm{\beta }\right)}^2+{\left(\tan \alpha -\tan \mathrm{\beta }\right)}^2={\sec }^2\alpha {\sec }^2\mathrm{\beta }$

Answer 1

${\left(1+\tan \alpha \tan \mathrm{\beta }\right)}^2+{\left(\tan \alpha -\tan \mathrm{\beta }\right)}^2={\sec }^2\alpha {\sec }^2\mathrm{\beta }$

$$\begin{gathered} \mathrm{LHS}={\left(1+\tan \alpha \tan \beta \right)}^2+{\left(\tan \alpha -\tan \beta \right)}^2 \\ =1+{\tan }^2\alpha {\tan }^2\beta +2\tan \alpha \tan \beta +{\tan }^2\alpha +{\tan }^2\beta -2\tan \alpha \tan \beta \\ =1+{\tan }^2\alpha {\tan }^2\beta +{\tan }^2\alpha +{\tan }^2\beta \\ ={\tan }^2\alpha \left({\tan }^2\beta +1\right)+1\left(1+{\tan }^2\beta \right) \\ =\left(1+{\tan }^2\beta \right)\left(1+{\tan }^2\alpha \right) \\ ={\sec }^2\mathrm{\alpha }.{\sec }^2\mathrm{\beta } \\ =\mathrm{RHS} \end{gathered}$$

Hence proved.