Solveeit Logo
Login

Question

Question: If \(\left( \frac { 1 + \sin \theta - \cos \theta } { 1 + \sin \theta + \cos \theta } \right) ^ { 2 ...

If (1+sin⁡θ−cos⁡θ1+sin⁡θ+cos⁡θ)2=λ(1−cos⁡θ1+cos⁡θ)\left( \frac { 1 + \sin \theta - \cos \theta } { 1 + \sin \theta + \cos \theta } \right) ^ { 2 } = \lambda \left( \frac { 1 - \cos \theta } { 1 + \cos \theta } \right), then λ equals

A

– 1

B

1

C

2

D

– 2

Answer

1

Explanation

Solution

(1+sin⁡θ−cos⁡θ1+sin⁡θ+cos⁡θ)2=(2sin⁡2θ2+2sin⁡θ2cos⁡θ22cos⁡2θ2+2sin⁡θ2cos⁡θ2)2\left( \frac { 1 + \sin \theta - \cos \theta } { 1 + \sin \theta + \cos \theta } \right) ^ { 2 } = \left( \frac { 2 \sin ^ { 2 } \frac { \theta } { 2 } + 2 \sin \frac { \theta } { 2 } \cos \frac { \theta } { 2 } } { 2 \cos ^ { 2 } \frac { \theta } { 2 } + 2 \sin \frac { \theta } { 2 } \cos \frac { \theta } { 2 } } \right) ^ { 2 }

= tan2 θ2\frac { \theta } { 2 } = 1−cos⁡θ1+cos⁡θ\frac { 1 - \cos \theta } { 1 + \cos \theta }

⇒ λ = 1