Question

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Answer 1

Approach :

Take sin θ + cos θ as common .

Use sin²θ + cos²θ = 1

Then finally 1/sin θ = cosec θ

And 1/cos θ = secθ

Given :

sin θ  [ sin θ + cos θ ] / cos θ + cos θ [ sin θ  + cos θ  ] / sin θ

Take sin θ  + cos θ  as common :

[ sin θ  + cos θ  ]  ( sin θ / cos θ  + cos θ / sin θ ]

= > [ sin θ  + cos θ  ] ( sin²θ  + cos² ) / sin θ cos θ

= > [ sin θ + cos θ ] / sin θ cos θ

= > 1 / cos θ + 1 / sin θ

= > sec θ + cosec θ

Hence Proved !

Answer 2

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