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Trignometry

Blazing goIITian

Joined:	20 Jul 2008
Post:	610

3 Jun 2009 19:58:28 IST

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PLZ try this .......................

Engineering Entrance , JEE Main , JEE Advanced , Mathematics , Trigonometry

Prove that

(tan4A + tan2A)(1 - 3a * A) = 2 * tan3A * A

Comments (2)

taran

Blazing goIITian

Joined: 17 Nov 2008

Posts: 1331

4 Jun 2009 14:19:17 IST

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tan 4A = (tan 3A + tanA)/(1- tan 3A tan A)

tan 2A=(tan 3A - tanA)/(1+ tan 3A tan A)

adding and rearranging

we get

(tan 4A + tan 2A) * ( 1 - tan ^ 2 A*tan ^2 3A)

= (tan 3A + tanA)(1+ tan 3A tan A) +(tan 3A - tanA) (1- tan 3A tan A)

further solving

LHS of question ==2 tan 3A + 2 (tan 3A)* (tan A)^2 (rest terms cancel out)

= 2 * tan3A * A

®µD®A

Blazing goIITian

Joined: 12 Apr 2008

Posts: 2717

4 Jun 2009 14:46:57 IST

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$http://latex.codecogs.com/gif.latex?(\tan%204A+\tan%202A)(1-\tan^22A\tan^2A)\\\\%20=\left(\frac{\sin4A}{\cos4A}+\frac{\sin2A}{\cos2A}\right)\left(1-\frac{\sin^23A\sin^22A}{\cos^23A\cos^2A}\right)\\\\%20=\left(\frac{\sin4A\cos2A+\cos4A\sin2A}{\cos4A\cos2A}\right)\left(\frac{\cos^2(3A)\cos^2(A)-\sin^2(3A)\sin^2(A)}{\cos^2(3A)\cos^2(A)}\right)\\\\%20=\frac{\sin%206A}{\cos%204A\cos%202A}\cdot\frac{\cos^23A\cos^22A-\sin^23A\sin^2A}{\cos^23A\cos^2A}\\\\%20=\frac{\sin%206A}{\cos%204A\cos%202A}\cdot\frac{(\cos%203A\cos%20A-\sin%203A\sin%20A)(\cos%203A\cos%20A+\sin%203A\sin%20A)}{\cos^23A\cos^2A}\\\\%20=\frac{2\sin%203A\cos%203A}{\cos%204A\cos%202A}\cdot\frac{\cos%204A\cos%202A}{\cos^23A\cos^2A}\\\\%20=2\tan%203A\cdot\frac{1}{\cos^2A}\\\\%20=2\tan%203A\sec^2A\\\\$

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