Question

Prove that cos theta-sin theta+1/ sin theta+cos theta-1 = 1 / cosec theta-cot theta​

Answer 1

Step-by-step explanation:

cosθ−sinθ+1

/* Divide numerator and denominator by sin\thetasinθ we get,

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

sinθ

1

(cosθ+sinθ−1)

sinθ

1

(cosθ−sinθ+1)

=\frac{(cot\theta-1+cosec\theta)}{(cot\theta+1-cosec\theta)}=

(cotθ+1−cosecθ)

(cotθ−1+cosecθ)

=\frac{(cot\theta+cosec\theta-1)}{(cot\theta+1-cosec\theta)}=

(cotθ+1−cosecθ)

(cotθ+cosecθ−1)

=\frac{(cot\theta+cosec\theta)-(cosec^{2}\theta-cot^{2}\theta)}{(cot\theta+1-cosec\theta)}=

(cotθ+1−cosecθ)

(cotθ+cosecθ)−(cosec

2

θ−cot

2

θ)

/* Since, By Trigonometric identity:

cosec²A-cot²A = 1 */

=\frac{(cot\theta+cosec\theta)-(cosc\theta-cot\theta)(cosec\theta+cot\theta)}{cot\theta-cosec\theta+1}=

cotθ−cosecθ+1

(cotθ+cosecθ)−(coscθ−cotθ)(cosecθ+cotθ)

/* By algebraic identity:

a²-b² = (a+b)(a-b) */

=\frac{(cosec\theta+cot\theta)[1-(cosec\theta-cot\theta)]}{(1-cosec\theta+cot\theta)}=

(1−cosecθ+cotθ)

(cosecθ+cotθ)[1−(cosecθ−cotθ)]

=\frac{(cosec\theta+cot\theta)(1-cosec\theta+cot\theta)}{(1-cosec\theta+cot\theta)}=

(1−cosecθ+cotθ)

(cosecθ+cotθ)(1−cosecθ+cotθ)

=cosec\theta+cot\theta=cosecθ+cotθ

=RHS=RHS

Therefore,

\frac{cos\theta-sin\theta+1}{cos\theta+sin\theta-1}=cosec\theta+cot\theta

cosθ+sinθ−1

cosθ−sinθ+1

=cosecθ+cotθ