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Trig Identity Question (1 Viewer)

TearsOfFire

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Prove the following identity

[sin A/(1 + cos A)] + [1 + cos A/(sin A)] = 2 cosec A

Thanks in advance
 
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LHS = (sin^2 A + 1 + 2cos A + cos^2 A) / [sin A(1 + cosA)]
LHS = (2 + 2cos A) / [sin A(1 + cos A)]
LHS = [2(1 + cos A)] / [sin A(1 + cos A)]
LHS = 2 / sin A
LHS = 2cosec A
LHS = RHS

hope this helped
 

randomnessss

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randomness you instead of subbing in 1-cos^2 A for sin^ A you could have just gone sin^2 A + cos^2 A = 1 its faster thats all. :D err i have to learn how to use this latex thingy now :l
lols k, thx bro - btw latex is annoying to use =(
 

Timothy.Siu

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[sin A/(1 + cos A)] + [1 + cos A/(sin A)] = 2 cosec A


LHS= sin A(1-cos A)/sin2 A + sin A(1+cos A)/sin2A
= 2sinA/sin2A=2cosec A
 

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