Find the value of (sin A + cosec A)² + (cos A + sec A)².

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Accepted Answer

Step 1: Write the expression:
(sin A + cosec A)² + (cos A + sec A)²

Step 2: Expand each square:
= (sin² A + 2 sin A · cosec A + cosec² A) + (cos² A + 2 cos A · sec A + sec² A)

Step 3: Use identities:
- sin A · cosec A = 1
- cos A · sec A = 1
- sin² A + cos² A = 1

So,
= (sin² A + 2 × 1 + cosec² A) + (cos² A + 2 × 1 + sec² A)
= (sin² A + cosec² A + 2) + (cos² A + sec² A + 2)
= (sin² A + cos² A) + (cosec² A + sec² A) + 4
= 1 + (cosec² A + sec² A) + 4
= 5 + cosec² A + sec² A

Step 4: Express cosec² A and sec² A in terms of tan² A and cot² A:
- cosec² A = 1 + cot² A
- sec² A = 1 + tan² A

So,
= 5 + (1 + cot² A) + (1 + tan² A)
= 5 + 1 + 1 + cot² A + tan² A
= 7 + tan² A + cot² A

Hence, the value is 7 + tan² A + cot² A.

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