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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

Blazing goIITian

Joined: 17 Nov 2008
Posts: 1331
4 Jun 2009 14:19:17 IST
4 people liked this

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

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

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

Blazing goIITian

Joined: 12 Apr 2008
Posts: 2717
4 Jun 2009 14:46:57 IST
2 people liked this

$(\tan 4A+\tan 2A)(1-\tan^22A\tan^2A)\\\\ =\left(\frac{\sin4A}{\cos4A}+\frac{\sin2A}{\cos2A}\right)\left(1-\frac{\sin^23A\sin^22A}{\cos^23A\cos^2A}\right)\\\\ =\left(\frac{\sin4A\cos2A+\cos4A\sin2A}{\cos4A\cos2A}\right)\left(\frac{\cos^2(3A)\cos^2(A)-\sin^2(3A)\sin^2(A)}{\cos^2(3A)\cos^2(A)}\right)\\\\ =\frac{\sin 6A}{\cos 4A\cos 2A}\cdot\frac{\cos^23A\cos^22A-\sin^23A\sin^2A}{\cos^23A\cos^2A}\\\\ =\frac{\sin 6A}{\cos 4A\cos 2A}\cdot\frac{(\cos 3A\cos A-\sin 3A\sin A)(\cos 3A\cos A+\sin 3A\sin A)}{\cos^23A\cos^2A}\\\\ =\frac{2\sin 3A\cos 3A}{\cos 4A\cos 2A}\cdot\frac{\cos 4A\cos 2A}{\cos^23A\cos^2A}\\\\ =2\tan 3A\cdot\frac{1}{\cos^2A}\\\\ =2\tan 3A\sec^2A\\\\$

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