Trigonometry : Simplify and find the value of $\tan\theta(1-\sec\frac{\theta}{2})(1-\sec\theta)(1-\sec2\theta)\dots(1-\sec2^{n-1}\theta)$ at n =1,2,3

Problem :

Simplify and find the value of $\tan\theta(1-\sec\frac{\theta}{2})(1-\sec\theta)(1-\sec2\theta)\dots(1-\sec2^{n-1}\theta)$ at n =1,2,3

My approach :

$\tan\theta(1-\sec\frac{\theta}{2})(1-\sec\theta)(1-\sec2\theta).....(1-\sec2^{n-1}\theta)$ ....(i)

$\tan\theta = \frac{2\tan\frac{\theta}{2}}{1-\tan^2\frac{\theta}{2}}$

=$\frac{2\sin\frac{\theta}{2} \cos\frac{\theta}{2}}{\cos^2\frac{\theta}{2}- \sin^2\frac{\theta}{2}}$

Putting this value in (i) we get :

$$= \frac{2\sin\frac{\theta}{2} \cos\frac{\theta}{2}}{\cos^2\frac{\theta}{2}- \sin^2\frac{\theta}{2}}(1-\sec\frac{\theta}{2})(1-\sec\theta)(1-\sec2\theta).....(1-\sec2^{n-1}\theta)$$

$$=\frac{2\sin\frac{\theta}{2}}{\cos^2\frac{\theta}{2}- \sin^2\frac{\theta}{2}}(\cos\frac{\theta}{2} - \sin\frac{\theta}{2} )(1-\sec\theta)(1-\sec2\theta).....(1-\sec2^{n-1}\theta)$$

$$=\frac{2\sin\frac{\theta}{2}}{\cos\frac{\theta}{2} + \sin \frac{\theta}{2}}(1-\sec\theta)(1-\sec2\theta).....(1-\sec2^{n-1}\theta)$$

Please suggest whether this is the right approach of solving this or we can use some other method.. thanks..

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$1-\sec(2^m\theta)=(1-\cos(2^m\theta))(\sec(2^m\theta)) =\frac{2\cos(2^{m-1}\theta)}{cos(2^m\theta)}$ – Math.StackExchange Oct 13 '13 at 16:44
Sorry, negative sign was lost: $1-\sec(2^m\theta)=-(1-\cos(2^m\theta))(\sec(2^m\theta)) =-\frac{2\cos(2^{m-1}\theta)}{cos(2^m\theta)}$ – Math.StackExchange Oct 13 '13 at 16:54

Use the following identity:

$1-\sec(2^m\theta)=-(1-\cos(2^m\theta))(\sec(2^m\theta)) =-\frac{2\cos(2^{m-1}\theta)}{cos(2^m\theta)}$

The given equation becomes:

$\tan(\theta)\frac{(-2)^{n+1}cos(\frac{1}{4}\theta)}{cos(2^{n-1}\theta)}$

If $n=1$, $4\cos(\theta/4)\tan(\theta)\sec(\theta)$

If $n=2$, $-8 \cos(\theta/4) \tan(\theta) \sec(2 \theta)$

If $n=3$, $16 \cos(\theta/4) \tan(\theta) \sec(4 \theta)$

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