Home » Ask & Discuss » Mathematics. » Trignometry « Back to Discussion

Trignometry

Blazing goIITian

Joined:	20 Aug 2010
Post:	342

4 Sep 2010 19:18:18 IST

0 People liked this

228

Ca this trigonometric identity be proved lhs = rhs

None

(1 + sinA)2 divided by (1-sinA)2 = 2cosA + sin A + Sin 3A divided by 2cosA - sin A - Sin 3A, i am not asking to prove it, i am asking that can this be proved if yes plz tell how!

Share this article on:

Comments (3)

rahulmishra

Blazing goIITian

Joined: 16 Jan 2010

Posts: 318

4 Sep 2010 21:35:26 IST

Like 1 people liked this


R.H.S = [2cosA + sinA + sin3A]/[2cosA - sinA - sin3A]

= [2cosA + sinA + 3sinA - 4sin³A]/'[2cosA - sinA - 3sinA + 4sin³A]
= [2cosA + 4sinA - 4sin³A]/[2cosA - 4sinA + 4sin³A]
= [2cosA + 4sinA(1 - sin²A)]/[2cosA - 4sinA(1 - sin²A)]
= [2cosA + 4sinA.cos²A]/[2cosA - 4sinA.cos²A]
= 2cosA [1 + 2sinA.cosA] / 2cosA [1 - 2sinA.cosA]
= (sinA + cosA)² / (sinA - cosA)²

But, L.H.S = (1 + sinA)² / (1 - sinA)² So i think there is some
error in the question. Please check!

Blazing goIITian

Joined: 20 Aug 2010

Posts: 342

5 Sep 2010 09:23:37 IST

Like 1 people liked this

i have already done this b4 posting the question bro

Elm

Cool goIITian

Joined: 30 Aug 2010

Posts: 45

7 Sep 2010 10:44:14 IST

Like 0 people liked this

well i too think that their is some mistake with the problem!

Quick Reply

Reply

	Some HTML allowed.
	Keep your comments above the belt or risk having them deleted.
	Signup for a avatar to have your pictures show up by your comment
	If Members see a thread that violates the Posting Rules, bring it to the attention of the Moderator Team