Question

Prove that ( 1 + 1/tan²A ) + ( 1 + 1/Cot²A ) = ( 1/Sin² A - Sin⁴A ).

Answer 1

Hey !!

LHS = ( 1 + 1/tan² A ) + ( 1 + 1/Cot²A )

= ( 1 + Cot²A ) ( 1 + Tan²A )

=> Cosec²A × Sec²A

=> 1/Sin²A × 1/Cos²A

=> 1/Sin²A Cos²A

=> 1/Sin²A ( 1 - Sin²A )

=> 1/ Sin²A - Sin⁴A ) = RHS

Hence,

LHS = RHS = 1/ Sin² A - Sin⁴A .

Answer 2

Heya!!

Here's your answer!!!
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The answer won't be equal to RHS it will be equal if the question is.

LHS
= (1+1/tan²A)(1+1/cot²A)

= ((tan²A+1)/tan²A)((cot²A+1)/cot²A)

= (sec²A/tan²A)(cosec²A/cot²A)

= (sec²A•cosec²A)/(tan²A•cot²A)

= sec²A• cosec²A

= 1/(sin²A•cos²A)

= (1/sin²A)(1/cos²A)

= (1/sin²A)(1/(1-sin²A))

= 1/(sin²-sin⁴•A)

Hence proved by LHS = RHS
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@vaibhav246