Question

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

Answer 1

Taking LHS=

2cos

2

θ

(1+sinθ)

2

(1−sinθ)

2

=

2cos

2

θ

1+sin

2

θ+2sinθ+1+sin

2

θ−2sinθ

[(a+b)

2

=a

2

+b

2

+2ab;(a−b)

2

=a

2

+b

2

−2ab]

=

2cos

2

θ

2+2sin

2

θ

=

2(1−sin

2

θ)

2(1+sin

2

θ)

[∵cos

2

θ+sin

2

θ=1]

=

(1−sin

2

θ)

(1+sin

2

θ)

=RHS

Answer 2

ans:- bro learn the identities you can solve easily