Question

1-cost heta÷1+cos theta= (cosec theta-cot theta)² prove that L.H.S=R.H.S​

Answer 1

Step-by-step explanation:

Given :-

(1- Cos θ)/(1+Cos θ)

To find :-

Prove that :

(1-Cos θ)/(1+Cos θ) = (Cosec θ- Cot θ)²

Solution :-

On taking LHS :

(1-Cos θ)/(1+Cos θ)

On multiplying both the numerator and the denominator with (1 - Cos θ) then

=>[(1-Cosθ)/(1+Cosθ)]×[(1-Cosθ)/(1-Cosθ)]

=>[(1- Cos θ)(1- Cos θ)]/[(1+ Cos θ)(1- Cos θ)]

=>[(1- Cos θ)²] / [(1+ Cos θ)(1- Cos θ)]

=> [(1- Cos θ)²] /[(1²- Cos² θ)]

Since (a+b)(a-b)=a²-b²

Where a = 1 and b = Cos θ

=> [(1- Cos θ)²] /( 1 - Cos² θ)

=> [(1- Cos θ)²] / Sin² θ

(Since Sin² θ + Cos² θ = 1

=> Sin² θ = 1-Cos² θ)

=> [(1- Cos θ)/ Sin θ]²

=> [ (1/Sin θ) - ( Cos θ/ Sin θ)]²

=> ( Cosec θ - Cot θ )²

=> RHS

=> LHS = RHS

Hence, Proved.

Answer:-

(1-Cos θ)/(1+Cos θ) = (Cosec θ- Cot θ)²

Used formulae:-

• (a+b)(a-b)=a²-b²

• Sin² θ + Cos² θ = 1

• Sin² θ = 1-Cos² θ

• Cosec θ = 1 / Sin θ

• Cot θ = Cos θ / Sin θ

Answer 2

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