Question

Plz solve this question....

Answer 1

Answer:

Step-by-step explanation:

2cos^2x+2/1+cot^2x

=2cos2x +2/cosec2x

=2cos^2x+2sin^2x

=2(cos^2x+sin^2x)

=2

Answer 2

To prove--->

2 Cos²θ + 2 / ( 1 + Cot²θ ) = 2

Proof--->

LHS = 2 Cos²θ + 2 / ( 1 + Cot²θ )

We have an identity, 1 + cot²A = Cosec²A , applying it here , we get,

= 2 Cos²θ + 2 / Cosec²θ

= 2 Cos²θ + 2 ( 1 / Cosec²θ )

= 2 Cos²θ + 2 ( 1 / Cosecθ )²

We know that , Sinθ = 1 / Cosecθ , applying it here , we get,

= 2 Cos²θ + 2 ( Sinθ )²

= 2 Cos²θ + 2 Sin²θ

Taking out 2 as common , we get,

= 2 ( Cos²θ + Sin²θ )

We know that , Cos²A + Sin²A = 1 , using it we get,

= 2 ( 1 )

= 2 = RHS