Trig Identity Question (1 Viewer)

[TearsOfFire]

Prove the following identity

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

Thanks in advance

 

[drowning_pocoyo]

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]

 

[drowning_pocoyo]

randomness instead of subbing in 1-cos^2 A for sin^2 A you could have just gone sin^2 A + cos^2 A = 1 its faster thats all. err i have to learn how to use this latex thingy now :l

 

[randomnessss]

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. 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]

[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