another Trig. sum

not sure i can kill this question... but just trying my hand at this...!!

setting 2^k = Θ we get,

tan (Θ/2).sec Θ = 1 - cos Θsin Θ (sec Θ) = 1sin Θ.co sΘ - 1sin Θ = 2sin 2Θ - 1sin Θ

so the expression turns into,

\sum_{k=1}^{n} \frac{2}{sin (2)^{k+1}} - \frac{1}{sin (2)^{k}}

= 2sin 4 - 1sin 2 + 2sin 8 - 1sin 4 + 2sin 16 - 1sin 8 + .................

= -1sin 2 + 1sin 4 + 1sin 8 + 1sin 16 + ............. + 1sin 2ⁿ + 2sin (2)^{n +1}

by using, 1sin 2Θ = cotΘ - cot2Θ

the expression can be furthur simplified to...

-1sin 2 + cot 2 - cot 2ⁿ + 2sin (2)^{n + 1}

= cos 2 - 1sin 2 + 2sin (2)^{n + 1} - cos 2ⁿsin 2ⁿ

= -tan1 + 2 - 2cos² (2)ⁿ2 sin 2ⁿcos 2ⁿ

= tan 2ⁿ - tan 1

does the answer match..!!

@man111 - is this result correct?

Yes Rahul fantastic Work man....

\hspace{-16}$Let $\mathbf{T_{k}=\tan (2)^{k-1}.\sec(2)^k=\frac{\sin(2^k-2^{k-1})}{\cos (2)^k.\cos(2)^{k-1}}}$\\\\\\ So $\mathbf{T_{k}=\tan(2)^k-\tan(2)^{k-1}}$\\\\\\ So $\mathbf{\sum_{k=1}^{n}T_{k}=\sum_{k=1}^{n}\{\tan(2)^k-\tan(2)^{k-1}\}}$\\\\\\ Now Using Telescoping Sum, We Get\\\\\\ So $\boxed{\boxed{\mathbf{\sum_{k=1}^{n}T_{k}=\tan(2)^n-\tan(1)}}}$