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Question Number 46007 by Meritguide1234 last updated on 19/Oct/18

	Answered by tanmay.chaudhury50@gmail.com last updated on 21/Oct/18
	$T_k =1−tan^4 ((π/2^k )) ={1−tan^2 ((π/2^k ))}{1+tan^2 ((π/2^k ))} ={((cos^2 ((π/2^k ))−sin^2 ((π/2^k )))/(cos^2 ((π/2^k ))))}×(1/(cos^2 ((π/2^k )))) T_k =((cos(((2π)/2^k )))/(cos^4 ((π/2^k ))))=((cos((π/2^(k−1) )))/({cos((π/2^k ))}^4 )) now T_3 ×T_4 ×T_5 ×T_6 ...×T_n =((cos((π/2^(3−1) ))cos((π/2^(4−1) ))cos((π/2^(5−1) ))...cos((π/2^(n−1) )))/([cos((π/2^3 ))cos((π/2^4 ))cos((π/2^5 ))...cos((π/2^n ))]^4 )) now see the first term of N_r =cos((π/2^2 )) see the nth of D_r =(1/([cos((π/2^n ))]^4 )) both remain unchanged... =((cos((π/2^2 )))/(cos^4 ((π/2^n ))))×(1/([cos((π/2^3 ))cos((π/2^4 ))cos((π/2^5 ))...cos((π/2^(n−1) ))]^3 )) =the red marked value to be calculated... wait... =((cos((π/4)))/(cos((π/2^n ))))×(1/([cos((π/2^3 ))cos((π/2^4 ))...cos((π/2^(n−1) ))cos((π/2^n ))]^3 )) sin((π/2^2 ))=2cos((π/2^3 ))sin((π/2^3 )) =2^2 cos((π/2^3 ))cos((π/2^4 ))sin((π/2^4 )) =2^3 cos((π/2^3 ))cos((π/2^4 ))cos((π/2^5 ))sin((π/2^5 )) cos((π/2^3 ))cos((π/2^4 ))cos((π/2^5 ))...cos((π/2^n ))=((sin((π/2^2 )))/(2^(f(n)) sin((π/2^n ))))=((sin((π/4)))/(2^(f(n)) ×sin((π/2^n )))) the value of f(n)to find...wait and condition n→∞ to put pls wait... now putting thevalue ((cos((π/4)))/(cos((π/2^n ))))×[((2^(f(n)) sin((π/2^n )))/(sin((π/4))))]^3 when n→∞ cos((π/2^n ))→1 but sin((π/2^n ))→0 so answer is zero pls check...$
	$T_{k} = 1 - {t a n}^{4} (\frac{π}{2^{k}})$ $= {1 - {t a n}^{2} (\frac{π}{2^{k}})} {1 + {t a n}^{2} (\frac{π}{2^{k}})}$ $= {\frac{{c o s}^{2} (\frac{π}{2^{k}}) - {s i n}^{2} (\frac{π}{2^{k}})}{{c o s}^{2} (\frac{π}{2^{k}})}} \times \frac{1}{{c o s}^{2} (\frac{π}{2^{k}})}$ $T_{k} = \frac{c o s (\frac{2 π}{2^{k}})}{{c o s}^{4} (\frac{π}{2^{k}})} = \frac{c o s (\frac{π}{2^{k - 1}})}{{c o s (\frac{π}{2^{k}})}^{4}}$ $n o w$ $T_{3} \times T_{4} \times T_{5} \times T_{6} . . . \times T_{n}$ $= \frac{c o s (\frac{π}{2^{3 - 1}}) c o s (\frac{π}{2^{4 - 1}}) c o s (\frac{π}{2^{5 - 1}}) . . . c o s (\frac{π}{2^{n - 1}})}{{[c o s (\frac{π}{2^{3}}) c o s (\frac{π}{2^{4}}) c o s (\frac{π}{2^{5}}) . . . c o s (\frac{π}{2^{n}})]}^{4}}$ $n o w s e e t h e f i r s t t e r m o f N_{r} = c o s (\frac{π}{2^{2}})$ $s e e t h e n t h o f D_{r} = \frac{1}{{[c o s (\frac{π}{2^{n}})]}^{4}}$ $b o t h r e m a i n u n c h a n g e d . . .$ $= \frac{c o s (\frac{π}{2^{2}})}{{c o s}^{4} (\frac{π}{2^{n}})} \times \frac{1}{{[c o s (\frac{π}{2^{3}}) c o s (\frac{π}{2^{4}}) c o s (\frac{π}{2^{5}}) . . . c o s (\frac{π}{2^{n - 1}})]}^{3}}$ $= t h e r e d m a r k e d v a l u e t o b e c a l c u l a t e d . . .$ $w a i t . . .$ $= \frac{c o s (\frac{π}{4})}{c o s (\frac{π}{2^{n}})} \times \frac{1}{{[c o s (\frac{π}{2^{3}}) c o s (\frac{π}{2^{4}}) . . . c o s (\frac{π}{2^{n - 1}}) c o s (\frac{π}{2^{n}})]}^{3}}$ $s i n (\frac{π}{2^{2}}) = 2 c o s (\frac{π}{2^{3}}) s i n (\frac{π}{2^{3}})$ $= 2^{2} c o s (\frac{π}{2^{3}}) c o s (\frac{π}{2^{4}}) s i n (\frac{π}{2^{4}})$ $= 2^{3} c o s (\frac{π}{2^{3}}) c o s (\frac{π}{2^{4}}) c o s (\frac{π}{2^{5}}) s i n (\frac{π}{2^{5}})$ $c o s (\frac{π}{2^{3}}) c o s (\frac{π}{2^{4}}) c o s (\frac{π}{2^{5}}) . . . c o s (\frac{π}{2^{n}}) = \frac{s i n (\frac{π}{2^{2}})}{2^{f (n)} s i n (\frac{π}{2^{n}})} = \frac{s i n (\frac{π}{4})}{2^{f (n)} \times s i n (\frac{π}{2^{n}})}$ $t h e v a l u e o f f (n) t o f i n d . . . w a i t a n d c o n d i t i o n$ $n \to \infty t o p u t p l s w a i t . . .$ $n o w p u t t i n g t h e v a l u e$ $\frac{c o s (\frac{π}{4})}{c o s (\frac{π}{2^{n}})} \times {[\frac{2^{f (n)} s i n (\frac{π}{2^{n}})}{s i n (\frac{π}{4})}]}^{3}$ $w h e n n \to \infty c o s (\frac{π}{2^{n}}) \to 1 b u t s i n (\frac{π}{2^{n}}) \to 0$ $s o a n s w e r i s z e r o p l s c h e c k . . .$
	Commented by Meritguide1234 last updated on 22/Oct/18

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