Math, asked by Dpadmavathidharani, 1 month ago

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Answered by tennetiraj86
1

Step-by-step explanation:

Given :-

√[(1+Sin θ)/(1-Sin θ)] - √[(1-Sin θ)/(1+Sin θ)]

To find :-

Prove that

√[(1+Sin θ)/(1-Sin θ)] -√[(1-Sin θ)/(1+Sin θ)] = 2/Cot θ

Solution :-

On taking LHS

√[(1+Sin θ)/(1-Sin θ)] -√[(1-Sinθ)/(1+Sinθ)]

On taking first part

√[(1+Sin θ)/(1-Sin θ)]

On multiplying both numerator and denominator with (1+Sin θ) then

=>√[{(1+Sin θ)/(1-Sin θ)}×[{(1+Sinθ)/(1+Sinθ)}]

=>√[{(1+Sin θ)(1+Sin θ)}×{(1-Sin θ)(1+Sin θ)}]

=>√[(1+Sin θ)²/{(1-Sin θ)(1+Sin θ)}]

=>√[(1+Sin θ)²/(1²-Sin² θ)]

Since , (a+b)(a-b) = a²-b²

Where, a = 1 , b = Sin θ

=> √[(1+Sin θ)²/(1-Sin² θ)]

=> √[(1+Sin θ)²/Cos² θ]

Since, Sin² A + Cos² A = 1

=> √[(1+Sin θ)/(Cos θ)]²

=> (1+Sin θ)/Cos θ

=> ( 1/Cos θ ) + ( Sin θ /Cos θ)

=> Sec θ + Tan θ ------------(1)

and

On taking second part

√[(1-Sin θ)/(1+Sin θ)]

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

=>√[{(1-Sinθ)/(1+Sinθ)}×[{(1-Sinθ)/(1-Sinθ)}]

=>√[{(1-Sin θ)(1-Sin θ)}×{(1+Sin θ)(1-Sin θ)}]

=>√[(1-Sin θ)²/{(1+Sin θ)(1-Sin θ)}]

=>√[(1-Sin θ)²/(1²-Sin² θ)]

Since , (a+b)(a-b) = a²-b²

Where, a = 1 , b = Sin θ

=> √[(1-Sin θ)²/(1-Sin² θ)]

=> √[(1-Sin θ)²/Cos² θ]

Since, Sin² A + Cos² A = 1

=> √[(1-Sin θ)/(Cos θ)]²

=> (1-Sin θ)/Cos θ

=> ( 1/Cos θ ) - ( Sin θ /Cos θ)

=> Sec θ - Tan θ ----------(2)

On Subtracting (2) from (1) then

=>√[(1+Sinθ)/(1-Sinθ)]-√[(1-Sinθ)/(1+Sinθ)]

=> (Sec θ + Tan θ) - ( Sec θ - Tan θ)

=> Sec θ + Tan θ - Sec θ +Tan θ

=> (Sec θ - Sec θ) +(Tan θ +Tan θ )

=> 0 + 2 Tan θ

=> 2 Tan θ

=> 2 ( 1/ Cot θ)

=> (2×1)/ Cot θ

=> 2 / Cot θ

=> RHS

=> LHS = RHS

Hence, Proved.

Answer :-

√[(1+Sin θ)/(1-Sin θ)] -√[(1-Sin θ)/(1+Sin θ)] = 2/Cot θ

Used formulae:-

→ (a+b)(a-b) = a²-b²

→ Sec θ = 1 / Cos θ

→ Tan θ = 1 / Cot θ

→ Sin θ / Cos θ = Tan θ

Answered by AkashMathematics
1

Here is the required answer in the attachment .

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