Question

Prove the following identities (1-17)

${\left(\sec A\sec B+\tan A\tan B\right)}^2-{\left(\sec A\tan B+\tan A\sec B\right)}^2=1$

Answer 1

$$\begin{gathered} \mathrm{LHS}={\left(\sec A\sec B+\tan A\tan B\right)}^2-{\left(\sec A\tan B+\tan A\sec B\right)}^2 \\ =\left[{\left(\sec A\sec B\right)}^2+{\left(\tan A\tan B\right)}^2-2\left(\sec A\sec B\right)\left(\tan A\tan B\right)\right] \\ -\left[{\left(\sec A\tan B\right)}^2+{\left(\tan A\sec B\right)}^2-2\left(\sec A\tan B\right)\left(\tan A\sec B\right)\right] \\ =\left[{\sec }^{2}A{\sec }^{2}B+{\tan }^{2}A{\tan }^{2}B-2\sec A\sec B\tan A\tan B\right] \\ -\left[{\sec }^{2}A{\tan }^{2}B+{\tan }^{2}A{\sec }^{2}B-2\sec A\sec B\tan A\tan B\right] \\ ={\sec }^{2}A{\sec }^{2}B+{\tan }^{2}A{\tan }^{2}B-2\sec A\sec B\tan A\tan B \\ -{\sec }^{2}A{\tan }^{2}B-{\tan }^{2}A{\sec }^{2}B+2\sec A\sec B\tan A\tan B \\ ={\sec }^{2}A{\sec }^{2}B-{\sec }^{2}A{\tan }^{2}B+{\tan }^{2}A{\tan }^{2}B-{\tan }^{2}A{\sec }^{2}B \\ ={\sec }^{2}A\left({\sec }^{2}B-{\tan }^{2}B\right)+{\tan }^{2}A\left({\tan }^{2}B-{\sec }^{2}B\right) \\ ={\sec }^{2}A\left({\sec }^{2}B-{\tan }^{2}B\right)-{\tan }^{2}A\left({\sec }^{2}B-{\tan }^{2}B\right) \\ ={\sec }^{2}A\times 1-{\tan }^{2}A\times 1 \\ ={\sec }^{2}A-{\tan }^{2}A \\ =1 \\ =\mathrm{RHS} \\ \mathrm{Hence}\mathrm{proved}. \end{gathered}$$