Question

(sin theta + sec theta) ^2+(cos theta+cosec theta) ^2=(1+cosec theta sec theta​

Answer 1

To prove:-

• (sinθ +secθ)² + (cosθ+cosecθ)²=(1+cosecθ.secθ)

Proof:-

LHS =  (sinθ +secθ)² + (cosθ+cosecθ)²

We know :

❥ Secθ = 1/cosθ

❥ Cosec θ = 1/sinθ

❥ Sin²θ + cos²θ = 1

⇒ (sinθ+1/cos θ)² + (cosθ+1/sinθ)²

⇒ (sinθcosθ+1 / cosθ )² + (sinθcosθ+1 / sinθ )²

⇒  (sinθcosθ+1 )² ( 1/ cos²θ  + 1/ sin²θ )

⇒ ( sin²θcos²θ + 2sinθcosθ +1 )[sin²θ + cos²θ / sin²θcos²θ ]

⇒  (sin²θcos²θ / sin²θcos²θ)+ (2sinθcosθ/sin²θcos²θ)+(1 /sin²θcos²θ )

⇒ 1 + 2 secθcosecθ + 2sec²θ cosec²θ

⇒  (1 + secθcosecθ)²

         = R.H.S.

HENCE PROVED !

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Answer 2

To prove:-

(sinθ +secθ)² + (cosθ+cosecθ)²=(1+cosecθ.secθ)

Proof:-

LHS =  (sinθ +secθ)² + (cosθ+cosecθ)²

We know :

❥ Secθ = 1/cosθ

❥ Cosec θ = 1/sinθ

❥ Sin²θ + cos²θ = 1

⇒ (sinθ+1/cos θ)² + (cosθ+1/sinθ)²

⇒ (sinθcosθ+1 / cosθ )² + (sinθcosθ+1 / sinθ )²

⇒  (sinθcosθ+1 )² ( 1/ cos²θ  + 1/ sin²θ )

⇒ ( sin²θcos²θ + 2sinθcosθ +1 )[sin²θ + cos²θ / sin²θcos²θ ]

⇒  (sin²θcos²θ / sin²θcos²θ)+ (2sinθcosθ/sin²θcos²θ)+(1 /sin²θcos²θ )

⇒ 1 + 2 secθcosecθ + 2sec²θ cosec²θ

⇒  (1 + secθcosecθ)²

         = R.H.S.

HENCE PROVED !

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