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Below is the trigonometric double angle problem. Please prove this.

If $\tan θ=\sec 2α$, prove that $\sin 2θ=\dfrac{1-\tan 4α}{1+\tan 4α}$

What I have tried:

R.H.S = $\dfrac{1-\tan 4α}{1+\tan 4α}$

= $\dfrac{cos4α-\sin4α}{cos4α+sin4α}$

= $\dfrac{cos^2(2α) - sin^2(2α) - 2sin2αcos2α}{cos^2(2α) - sin^2(2α) + 2sin2αcos2α}$

.....

......

Just the hit will do.

Sorry everybody, it was a false proof, just got the wrong questions. It was supposed to be

$\sin 2θ=\dfrac{1-\tan^4α}{1+\tan^4α}$

And i have solved it already, thanks everyone for valuable comments.

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3  
..and you have done what so far? –  DonAntonio Dec 15 '13 at 17:20
    
I tried to prove from the RHS, converting tan4α to sin4α/cos4α and then solving and applying double angle formula on cos4α with cos2.2α and then apply 1/cos2α=tanθ.... –  user065 Dec 15 '13 at 17:23
    
Then you should add that to your question, perhaps under "I"ve tried the following:" or like that, otherwise most members will think you're tying others to do your homework for you and, besides downvoting you, perhaps you'll get your question closed. –  DonAntonio Dec 15 '13 at 17:24
    
@DonAntonio can you just provide the hint so that I can prove it myself –  user065 Dec 15 '13 at 17:34
1  
Haven't tried this myself but it might be worth using the half-angle identity: $$\sin(2\theta)=\frac{2\tan(\theta)}{1+\tan^2(\theta)}$$ –  Mufasa Dec 15 '13 at 17:47

1 Answer 1

The claim is false.

Choose $\theta=\arctan (-2)$ and $\alpha=\dfrac{\pi}{3}$. Then $\tan \theta=\sec 2\alpha$.

But $\sin 2\theta=2\sin\theta\cos\theta=-2\cdot\sqrt{\dfrac{\tan^2\theta}{1+\tan^2\theta}}\cdot\sqrt{\dfrac1{1+\tan^2\theta}}=-2\cdot\sqrt{\dfrac45}\cdot\sqrt{\dfrac15} =-0.8$

and $\dfrac{1-\tan 4\alpha}{1+\tan 4\alpha}=\dfrac{1-\sqrt3}{1+\sqrt3}=2-\sqrt3\not=-0.8.$

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ya you are right. I got the wrong question. It was supposed to be tan to fourth power of alpha. Thanks –  user065 Dec 18 '13 at 16:14

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