Question

33. Prove that (Sin A + Cosec A) + (Cos A +Sec A) = 7 + tanA+ Cot A​

Answer 1

Question:

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

Answer:

Step-by-step explanation:

Given:

• (sin A + cosec A)² + (cos A + sec A)² = 7 + tan² A + cot² A

To Prove:

LHS = RHS

Identities used:

(a + b)² = a² + 2ab + b²

sin A × cosec A = 1

cos A × sec A = 1

sin² A + cos² A = 1

sec² A = 1 + tan² A

cosec² A = 1 + cot² A

Proof:

Taking the LHS of the equation,

LHS = (sin A + cosec A)² + (cos A + sec A)²

Expanding by applying the identities

⇒ sin² A + 2sin A cosec A + cosec² A + cos²A  + 2cos A secA + sec² A

Again using suitable identities,

⇒ sin² A + 2 × 1 + cosec² A + cos² A + 2 × 1 + sec² A

⇒ 2 + 2 + sin² A + cos²A + cosec² A + sec² A

⇒ 4 + 1 + cosec² A + sec² A

⇒ 5 + cosec² A + sec² A

Now we know that,

sec² A = 1 + tan² A

cosec² A = 1 + cot² A

Hence,

⇒ 5 + 1 + cot² A + 1 + tan² A

⇒ 7 + tan² A + cot² A

= RHS

Hence proved.