Trigonometric Identities

[sushifam]

Prove LS = RS

(sinθ)/1+cosθ =1/sinθ -1/tanθ

 

[R.M.]

And you have tried......?

 

[Deleted member 4993]

Prove LS = RS

(sinθ)/1+cosθ =1/sinθ -1/tanθ

Your problem statement should be:

(sinθ)/(1+cosθ) = 1/sinθ -1/tanθ ....... Those parentheses () are super important.

If I were to solve this problem, I would "attack" the LS and determine:

the conjugate of the denominator \(\displaystyle \ \ \to \ \ \)1 - cosθ .​

Please show us what you have tried and exactly where you are stuck.

Please follow the rules of posting in this forum, as enunciated at:

READ BEFORE POSTING​

Please share your work/thoughts about this problem.

 

[pka]

Prove LS = RS
(sinθ)/1+cosθ =1/sinθ -1/tanθ

I would note that \(\dfrac{1}{tan(\theta)}=\dfrac{\cos(\theta)}{\sin(\theta)}\)
That gives a common denominator on the RHS.

 

[Steven G]

And you have tried......?

Finally a question I can answer!! But I'll be nice and not answer it here.