Question

Prove the following.

(cos θ − sin θ) ( 1 + tan θ)/

2cos

θ − 1

= sec θ​

Answer 1

Answer:

To Prove:

tanθ−secθ+1tanθ+secθ−1=cosθ1+sinθ

Solution:

L.H.S =tanθ−secθ+1tanθ+secθ−1

We can write, sec2θ−tan2θ=1

=tanθ−secθ+1tanθ+secθ−(sec2θ−tan2θ)

=tanθ−secθ+1tanθ+secθ−(secθ−tanθ)(secθ+tanθ)

=tanθ−secθ+1(tanθ+secθ){1−(secθ−tanθ)}

=tanθ−secθ+1(tanθ+secθ){1−secθ+tanθ}

=tanθ+secθ

=cosθsinθ+cosθ1

hence proved