Prove that Cos A/1- sin A + Cos A/ 1+ sin A= 2sec A...good answer will mark as brainlist
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Prove that Cos A/1- sin A + Cos A/ 1+ sin A= 2sec A...good answer will mark as brainlist
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Step-by-step explanation:
Given :-
Cos A/(1- sin A) + Cos A/( 1+ sin A)
To Find :-
Cos A/(1- sin A) + Cos A/ (1+ sin A) = 2sec A.
Solution :-
On taking LHS in the equation
Cos A/(1- sin A) + Cos A/ (1+ sin A)
=> Cos A [ 1/(1-Sin A) + 1/(1+Sin A)]
=>CosA[{(1+SinA)+(1-SinA)}/(1-SinA)(1+SinA)]
=> Cos A [ (1+Sin A +1-Sin A)/(1²-Sin²A)]
Since (a+b)(a-b) = a²-b²
Where, a = 1 and b = Sin A
=> Cos A [ (1+1)/(1-Sin² A)]
=> Cos A [ 2/(1-Sin² A)]
We know that
Sin² A + Cos² A = 1
=> Cos A (2/Cos² A)
=> 2Cos A / Cos² A
=> 2 Cos A /(Cos A × Cos A)
On cancelling Cos A in both the numerator and the denominator then
=> 2/Cos A
=> 2(1/Cos A)
=> 2 Sec A
=> RHS
=> LHS = RHS
Hence, Proved.
Answer :-
Cos A/(1- sin A) + Cos A/(1+ sin A) =2secA.
Used Identities :-
→ (a+b)(a-b) = a²-b²
→ Sin² A + Cos² A = 1
→ 1/Cos A = Sec A
Step-by-step explanation:
[tex]\frac{cosA}{1-sinA} +\frac{cosA}{1+sinA} = cosA[ \frac{1}{1-sinA} +\frac{1}{1+sinA}]\\\\=cosA\frac{(1+sinA)+(1-sinA)}{(1+sinA)(1-sinA)} \\\\=\frac{2cosA}{1-sin^2A} \\\\=\frac{2cosA}{cos^2A}\\\\\=\frac{2}{cosA}\\\\=2secA[/tex]