sin teta /1+cos tera + 1+ cos teta/sin teta =1+cosec teta
Answers
Answer:
cosθ+sinθ−1
cosθ−sinθ+1
=cosecθ+cotθ
Step-by-step explanation:
\frac{cos\theta-sin\theta+1}{cos\theta+sin\theta-1}
cosθ+sinθ−1
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
