sinA÷1+cosA=?
(a)1+cosA÷sinA. (b)1-cosA÷cosA
Answers
Answer:
1
2
What is the ans for:(1-cosA+sinA/1+cosA +sinA) ^2=1-cosA/1+cos A?
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Gareth Morgan
Answered March 4, 2019
On the left side of the equation
Write CosA ( numerator) as 1–(2 sin^2 {A/2})
Write Cos A (denominator) as (2 cos^2{A/2})-1
Write Sin A as {2 Sin(A/2) Cos (A/2)}
We get
1 - [1–(2 sin^2 {A/2})] + 2 Sin(A/2) Cos(A/2) whole squared in the numerator
And
1 + [(2 cos^2{A/2})-1]+ 2 Sin(A/2) Cos(A/2) whole squared in the denominator
[2 Sin (A/2) {Sin(A/2)+ Cos(A/2)}/ 2 Cos (A/2){Sin(A/2)+ Cos(A/2)}]^2
Cancelling out the common parts - [2{Sin(A/2)+ Cos(A/2)}]^2
We get tan^2 (A/2)
Write it as Sin^2(A/2)/Cos^2(A/2)
Multiply by 2 on the numerator and denominator
Add and subtract 1, on the numerator and denominator
1–1+
Ved Prakash Sharma
Answered March 23, 2019
{(1-cosA+sinA)/(1+cosA+sinA)}^2= (1-cosA)/(1+cosA)
L.H.S.
={(1-cosA)+sinA}^2/{(1+cosA)+sinA}^2.
={(1-cosA)^2+sin^2A+2.(1-cosA).sinA}/{(1+cosA)^2+sin^2A+2.(1+cosA).sinA}.
={1–2.cosA+cos^2A+sin^2A+2.(1-cosA).sinA}/{1+2.cosA+cos^2A+sin^2A+2.
(1+cosA).sinA}.
={2(1-cosA)+2.(1-cosA).sinA}/{2(1+cosA)+2.(1+cosA).sinA}
= 2.(1-cosA).(1+sinA)/2.(1+cosA).(1+sinA).
= (1-cosA)/(1+cosA). Proved.
Answer:
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