Question

1 + cos theta / 1 - cos theta =( cosec theta + cot theta )²

*PROVE LHS = RHS*​

Answer 1

Step-by-step explanation:

Given :-

(1+Cos θ) / (1-Cos θ)

To Prove :-

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

Proof :-

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+Cosθ)(1+Cosθ)]/[(1-Cosθ)(1+Cosθ)]

=> (1+Cos θ)² / [(1-Cos θ)(1+Cos θ)]

=> (1+Cos θ)² / [1²-(Cos θ)²]

Since , (a+b)(a-b) = a²-b²

Where, a = 1 and b = Cos θ

=> (1+Cos θ)² / [1-Cos² θ]

We know that

Sin² θ + Cos² θ = 1

So,

=> [ (1+Cos θ)² / Sin² θ

=> [ (1+Cos θ)² / Sin² θ

=> [(1+Cos θ)/ Sin θ]²

=> [(1/Sin θ) + ( Cos θ/Sin θ)]²

=> (Cosec θ + Cot θ)²

Since , 1/Sin θ = Cosec θ

and Cot θ = Cos θ / Sin θ

=> RHS

=> LHS = RHS

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

Hence, Proved.

Used formulae :-

→ 1/Sin θ = Cosec θ

→ Cot θ = Cos θ / Sin θ

→ (a/b)^m = a^m / b^m

Used Trigonometric Identities :-

→ Sin² θ + Cos² θ = 1

Used Algebraic Identities:-

→ (a+b)(a-b) = a²-b²