Prove that \((sin A + cosec A)^2 + (cos A + sec A)^2 = 7 + tan^2A + cot^2A\).
Proof:
\(LHS=(sin A + cosec A)^2 + (cos A + sec A)^2\)
We know that \(cosec A = \frac{1}{sin A}\)
Therefore,
= \(7 + tan^2A + cot^2A\)
\(= RHS\)
Answer variants:
\(= sin^2A + cosec^2A + 2sin Acosec A + cos^2A + sec^2A + 2cos A\) \(sec A\)
\(LHS= 1 + 1 + cot^2A + 1 + tan^2A + 2 + 2 \)