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Answers
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
Here is the required answer in the attachment .
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![](https://hi-static.z-dn.net/files/da4/53aac241d48df49d910d230dc5746803.jpg)
![](https://hi-static.z-dn.net/files/db1/7fabacc120c051c69ed72a9db65906b0.jpg)
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![](https://hi-static.z-dn.net/files/deb/93dec90ff92abb8b721639f8958eec2a.jpg)