Question

(1/cos theta + sin theta/cos theta) ( 1- sin theta/cos theta)​

Answer 1

Answer:

1+cosθ+sinθ

=

cosθ

1+sinθ

L.H.S.

\frac{1+cos\theta + sin\theta}{1+cos\theta - sin\theta}

1+cosθ−sinθ

1+cosθ+sinθ

=\frac{1+cos\theta + sin\theta}{1+cos\theta - sin\theta}\times \frac{1+cos\theta + sin\theta}{1+cos\theta + sin\theta}=

1+cosθ−sinθ

1+cosθ+sinθ

×

1+cosθ+sinθ

1+cosθ+sinθ

=\frac{(1+cos\theta)^2+sin^2\theta + 2 sin\theta(1+cos\theta)}{1+cos^2\theta + 2cos\theta - sin^2 \theta}=

1+cos

2

θ+2cosθ−sin

2

θ

(1+cosθ)

2

+sin

2

θ+2sinθ(1+cosθ)

=\frac{1+cos^2\theta + 2 cos\theta + sin^2\theta + 2 sin\theta(1+cos\theta)}{1+cos^2\theta + 2cos \theta - 1 + cos^2 \theta}=

1+cos

2

θ+2cosθ−1+cos

2

θ

1+cos

2

θ+2cosθ+sin

2

θ+2sinθ(1+cosθ)

=\frac{1+1+2cos \theta + 2 sin \theta( 1+cos\theta)}{2cos^2\theta + 2 cos\theta}=

2cos

2

θ+2cosθ

1+1+2cosθ+2sinθ(1+cosθ)

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

2

θ+cos

2

θ=1)

=\frac{2+2cos\theta + 2sin\theta cos\theta + 2 sin \theta}{2cos^2 \theta + 2 cos \theta }=

2cos

2

θ+2cosθ

2+2cosθ+2sinθcosθ+2sinθ

=\frac{2(1+cos\theta)+2sin \theta ( 1+ cos \theta)}{2cos \theta (cos \theta + 1)}=

2cosθ(cosθ+1)

2(1+cosθ)+2sinθ(1+cosθ)

=\frac{1+sin\theta }{cos \theta }=

cosθ

1+sinθ

R.H.S.