Question

Attention!!!


Prove it :


(CosecØ-cotØ)² = (1-cosØ)/(1+cosØ)​

Answer 1

$\color{blue}\huge\underline\mathfrak{Solution:-}$

Taking LHS,

(CosecØ-cotØ)²

=> [(1/sinØ) - (cosØ/sinØ)]²

=> [(1-cosØ)/sinØ]²

=> (1-cosØ)²/sin²Ø

Taking RHS,

(1-cosØ)/(1+cosØ)

Rationalise it :

=> [(1-cosØ)(1-cosØ)]\[(1+cosØ)(1-cosØ)

=> (1-cosØ)²/(1-cos²Ø) [ a²-b² = (a+b)(a-b) ]

=> (1-cosØ)²/sin²Ø

LHS = RHS

Hence proved!

___________________________

$\large\underline\bold\red{Identity\:used:-}$

★ 1 - cos²Ø = sin²Ø

$\large\underline\bold\red{Trigonometrical\:Ratios\:used:-}$

★ cosecØ = 1/sinØ

★ cotØ = cosØ/sinØ

Answer 2

Answer:

Proved.

Step-by-step explanation:

Given -

(CosecØ-cotØ)² = (1-cosØ)/(1+cosØ)​

To Prove -

(CosecØ-cotØ)² = (1-cosØ)/(1+cosØ)​

Proof -

Firstly we have to take LHS & RHS.

So,

Taking LHS -

(CosecØ-cotØ)²

Now,

$\sf\\ \\\implies \left[\dfrac{1}{sin\theta}- \dfrac{cos\theta}{sin\theta^2}\right]\\ \\ \\ \implies \dfrac{1-cos\theta}{sin\theta^2}\\ \\ \\ \implies \dfrac{1-cos\theta^2}{sin^2\theta}$

Now,

Taking RHS,

$\sf \dfrac{1-cos\theta}{1+cos\theta}$

__________{ Rationalising it. }

$\sf \\ \\\implies \left[\dfrac{(1-cos\theta)(1-cos\theta)}{(1+cos\theta)(1-cos\theta)}\right]\\ \\ \\\implies \dfrac{(1-cos\theta)^2}{(1-cos^2\theta)}\quad[(a+b)(a-b)=a^2-b^2]\\ \\ \\\implies \dfrac{1-cos\theta^2}{sin^2\theta}$

LHS = RHS.

Hence, Proved.

$\rule{300}{1.5}$

★ Identity Used -

⇒ 1 - cos²Ø = sin²Ø