Question

Solve 2 cos 4 tita cos 2 tita​

Answer 1

Answer:

Taking LHS=2cos

2

θ−cos

4

θ+sin

4

θ=2cos

2

θ−(cos

4

θ−sin

4

θ)=2cos

2

θ−[(cos

2

θ)

2

−(sin

2

θ)

2

]

using identity (a

2

−b

2

)=(a+b)(a−b)

=cos

2

θ−[(cos

2

θ−sin

2

θ)(cos

2

θ+sin

2

θ)]=2cos

2

θ−[(cos

2

θ−sin

2

θ)(1)][∵cos

2

θ+sin

2

θ=1]

=2cos

2

θ−cos

2

θ+sin

2

θ=cos

2

θ+sin

2

θ=1=RHS