prove that 1-sin2x/1+sin2x=tan^2(pi/4-x)
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(1-sin2x)/(1+sin2x)
=(sin90-sin2x)/(sin90+sin2x)
={2cos(90+2x)/2.sin(90-2x)/2}/2sin(90+2x)/2.cos(90-2x)/2
=cot(45+x).tan(45-x)
=cot{90-(45-x)}.tan(45-x)
=tan^2(45-x) =RHS
=(sin90-sin2x)/(sin90+sin2x)
={2cos(90+2x)/2.sin(90-2x)/2}/2sin(90+2x)/2.cos(90-2x)/2
=cot(45+x).tan(45-x)
=cot{90-(45-x)}.tan(45-x)
=tan^2(45-x) =RHS
harivermasharma:
by the way, which guide do you refer to for math!?
no any this is my choice ?
please mark as brainliest
I can't see any such option on my screen.
actually i'm new to this
okay no problem
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