Prove that (sin A + cosec A)² + (cos A + sec A)² = 7 + tan² A + cot² A…

Step 1. Expand. The left side is $\sin^2 A + 2 + \text{cosec}^2 A + \cos^2 A + 2 + \sec^2 A$.

Step 2. Group. $(\sin^2 A + \cos^2 A) + 4 + \text{cosec}^2 A + \sec^2 A = 1 + 4 + (1 + \cot^2 A) + (1 + \tan^2 A)$.

Step 3. Add up. That is $7 + \tan^2 A + \cot^2 A$, the right side.

Step 4. Check at 45°. The right side is $7 + 1 + 1 = 9$. On the left, $\left(\dfrac{1}{\sqrt{2}} + \sqrt{2}\right)^2 = \dfrac{9}{2}$, twice over, which is $9$.