Question

Prove that

(Sin A + Cos A) ² + (Cos A + Sec A) ² = 7 + Tan² A + Cot² A​

Answer 1

Correct Question:

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

Here, Identity used,

• 1 + tan² A = sec²A
• 1 + cot² A = cosec² A
• sin² A + cos² A = 1

Also,

• Cosec A = 1/sinA
• sec A = 1/cos A

Proof:

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

Taking LHS,

⇒ (sin A + cosec A)² + (cos A + sec A)²

⇒ sin² A + cosec² A + 2 sin A cosec A + cos² A + sec² A + 2 cos A sec A [ (a + b)² = a² + b² + 2ab ]

⇒ sin² A + cosec² A + 2 sin A × 1/sin A + cos² A + sec² A + 2 cos A × 1/cos A

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

⇒ 1 + cosec² A + 2 + sec² A + 2 [ sin² A + cos² A = 1 ]

⇒ 5 + cosec² A + sec² A

⇒ 5 + (1 + cot² A) + (1 + tan ² A)

⇒ 7 + tan² A + cot² A

★ LHS = RHS

Hence, Proved!

Answer 2

Answer:

Here, Identity used,

1+ tan? A = sec²A

1+ cot? A = cosec? A sin? A + cos? A = 1

Cosec A = 1/sinA

• sec A = 1/cos A =

Proof:

(sin A + cosec A)2 + (cos A + sec A)? = 7 + tan? A + cot? A

Taking LHS,

→ (sin A + cosec A)2 + (cos A + sec

A)2

→ sin? A + cosec? A + 2 sin A cosec

A + cos? A + sec? A + 2 cos A sec A

[ (a + b)? = a² + b2 + 2ab ] =

sin? A + cosec? A + 2 sin A x 1/sin

A + cos? A + sec? A + 2 cos A x 1/cos

A

sin? A + cosec² A + 2 + cos? A +

sec? A + 2

+ 1 + cosec? A + 2 + sec? A + 2

[ sin? A + cos?A =1]

- 5 + cosec? A + sec? A

- 5 + (1+ cot? A) + (1 + tan 2 A)

→ 7 + tan? A + cot? A

* LHS = RHS

HENCE PROVED