Question

Prove.

1/ (cosecΦ - cot Φ) - 1/ sinΦ = 1/sinΦ - 1/ (cosecΦ + cot Φ)​

Answer 1

Step-by-step explanation:

L.H.S → cot²∅/(cosec∅ - 1)

L.H.S → cot²∅/(1/sin∅ - 1)

L.H.S → (cos∅/sin∅)²/(1 - sin∅)/sin∅

L.H.S → cos²∅/sin²∅ × sin∅/(1 - sin∅)

L.H.S → cos²∅/sin∅ × 1/(1 - sin∅)

L.H.S → (1 - sin²∅)/sin∅(1 - sin∅)

L.H.S → (1 + sin∅)(1 - sin∅)/sin∅(1 - sin∅)

L.H.S → (1 + sin∅)/sin∅ = R.H.S

hope will be use

Answer 2

Given :

• LHS :

$\mathtt{\dfrac{1}{(cosec\:\theta\:-\:cot\:\theta)}}$ - $\mathtt{\dfrac{1}{sin\:\theta}}$

Given :

• RHS :

$\mathtt{\dfrac{1}{sin\:\theta}\:-\:\dfrac{1}{sin\:\theta}\:-\:\dfrac{1}{(cosec\:\theta\:+\:cot\:\theta}}$

Proof :

Taking LHS :

$\mathtt{\dfrac{1}{(cosec\:\theta\:-\:cot\:\theta)}}$ - $\mathtt{\dfrac{1}{sin\:\theta}}$

$\mathtt{\dfrac{1}{(cosec\:\theta\:-\:cot\:\theta)}\:\times\:{\dfrac{(cosec\:\theta\:+\:cot\:\theta)}{(cosec\:\theta\:+\:cot\:\theta)}\:-\:cosec\:\theta}}$

$\mathtt{\dfrac{(cosec\:\theta\:+\:cot\:\theta)}{(cosec^2\:\theta\:-\:cot^2\:\theta)}\:-\:cosec\:\theta}$ $\mathtt{\underbrace{\Big[\because\:\dfrac{1}{sin\:\theta}\:=\:cosec\:\theta\Big]}}$

$\mathtt{(cosec\:\theta\:+\:cot\:\theta)\:-\:(cosec\:\theta)}$ $\mathtt{\underbrace{\Big[\because\:cosec^2\:\theta\:-cot^2\theta\:=\:1\Big]}}$

$\mathtt{\cancel{cosec\theta}\:+\:cot\:{\cancel{-cosec\:\theta}}}$

$\large{\boxed{\tt{\red{ LHS\:=\:cot\:\theta}}}}$

Now, taking RHS :

$\mathtt{\dfrac{1}{sin\:\theta}\:-\:\dfrac{1}{sin\:\theta}\:-\:\dfrac{1}{(cosec\:\theta\:+\:cot\:\theta)}}$

$\mathtt{cosec\:\theta\:-\:\dfrac{1}{(cosec\:\theta\:+\:cot\:\theta)}\:\times\:\dfrac{(cosec\theta\:-\:cot\:\theta)}{(cosec\:\theta\:-\:cot\:\theta)}}$

$\mathtt{cosec\:\theta\:-\:\dfrac{(cosec\theta\:-\:cot\:\theta)}{(cosec^2\:\theta\:-\:cot^2\:\theta)}}$

$\mathtt{cosec\:\theta\:-\:(cosec\:\theta\:-\:cot\:\theta)}$ $\mathtt{\underbrace{\Big[\because\:cosec^2\theta\:-\:cot^2\theta\:=1\:\Big]}}$

$\mathtt{cosec\:\theta\:-\:cosec\:\theta\:+cot\:\theta}$

$\mathtt{\cancel{cosec\:\theta}\:{\cancel{-cosec\:\theta}+\:cot\:\theta}}$

$\large{\boxed{\tt{\red{ RHS\:=\:cot\:\theta}}}}$

•°• LHS = RHS.