Question

$prove \: that \: ( cosec \: theta - \cot \: theta {)}^{2} = \binom{1 - \cos \: theta }{1 + \cos \: theta}$
please its tooo urgent
Answer me faster​

Answer 1

Step-by-step explanation:

Given :-

(Cosec θ - Cot θ)²

To find :-

Prove that

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

Solution :-

On taking LHS

(Cosec θ - Cot θ)²

=> [ (1/Sin θ) - (Cos θ/ Sin θ) ] ²

LCM = Sin θ

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

On taking RHS

=> ( 1-Cos θ)/(1+Cos θ)

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

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

=> (1-Cos θ)²/(1²-Cos² θ)

=> (1-Cos θ)²/(1-Cos² θ)

=> (1-Cos θ)²/Sin² θ

Since Sin² A + Cos² A = 1

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

From (1)&(2)

LHS = RHS

Therefore,

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

Hence, Proved.

Used formulae:-

→ Sin² A + Cos² A = 1

→ Cosec A = 1/SinA

→ Cot A = Cos A / Sin A

→ (a/b)^m = a^m / b^m

Answer 2

Step-by-step explanation:

(cosecA-cotA)^2=1-cosA÷1+cosA

(1/sinA-cosA/sinA)^2

(1-cosA÷sinA)^2 (1-cosA)^2-sin^2A

(1-cosA)(1-cosA)÷1-cos^2A

(1-cosA)(1-cosA)÷(1-cosA)(1+cosA) (1-cosA)=(1+cosA) proved