Question

show that tan theta(1+sec 2 theta)(1+sec 2^2 theta)(1+sec 2^3 theta)...(1+sec 2^n theta) =tan 2^n theta. ​

Answer 1

Step-by-step explanation:

Note that

$1 + sec \: 2x \\ \\ =1 + \frac{1 + tan {}^{2} \: x}{1 - tan {}^{2}x } \\ \\ = \frac{2}{1 - {tan}^{2}x } \\ \\ = \frac{1}{tan \: x} \frac{2tan \: x}{1 - {tan}^{2}x } \\ \\ = cot \: x \: tan \: 2x$

Which will result a recursive formula

$= (cot \: x \: tan \: 2x)(cot 2x \tan \: 4x)....(cot \: {2}^{n - 1} x \: tan {2}^{n} x)$

So All terms of ( tan and cot of same angle ) will cancel out except first and last

$= \cot(x) \tan( {2}^{n}x )$

Answer 2

Step-by-step explanation:

solution :

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