Math, asked by rampirkash11, 6 months ago

sinA=(a-b/a+b) than proove that tan(pi/4-A/2)=+-√b/a

Answers

Answered by Ayodhyavasi

solution:

SinA= (a-b)/(a+b)

1/SinA=(a+b)/(a-b).........(i)

by componendo and devidendo

we have

(1+SinA)/(1-SinA)=(a+b+a-b)/(a+b-a+b).......(ii)

(1+sinA)/(1-sinA) = 2a/2b= a/b

since we can write

SinA=Cos(π/2-A)

=Cos(π-2A)/2

and

CosA= 2cos^2(A/2) -1

= 1- 2sin^2(A/2)

then

1+SinA=1+Cos(π-2A)/2

= 1+ 2cos^2(π/4-A/2)-1

=2cos^2(π/4-A/2)

Similarly

1-SinA = 2Sin^2(π/4-A/2)

from equation (ii)

we have

2cos^2(π/4-A/2)/2sin^2(π/4-A/2) = a/b

cos^2(π/4-A/2)/ Sin^2(π/4-A/2) = a/b

Cot^2(π/ 4-A/2) = a/b

tan^2(π/4-A/2) = b/a

tan(π/ 4-A/2) = +-√b/ a

Proved.

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sinA=(a-b/a+b) than proove that tan(pi/4-A/2)=+-√b/a