Question

Prove that

$(1+\sec 2\theta )(1+\sec 4\theta )(1+\sec 8\theta )....(1+\sec 2^n\theta )=\tan \left(2^n\theta \right).\cot \theta$.

Answer 1

L.H.S

$\Rightarrow (1+sec2\theta )(1+sec4\theta )(1+sec8\theta )...(1+sec{2}^n\theta )$

$\Rightarrow (1+\frac{1}{cos2\theta })(1+sec4\theta )(1+sec8\theta )...(1+sec{2}^n\theta )$

$\Rightarrow (1+\frac{1+ta{n}^2\theta }{1-ta{n}^2\theta })(1+sec4\theta )(1+sec8\theta )...(1+sec{2}^n\theta )$

$\Rightarrow \left(\frac{2}{1-ta{n}^2\theta }\right)(1+sec4\theta )(1+sec8\theta )...(1+sec{2}^n\theta )$

Multiply with $tan\theta$ & also Divides the same.

$\Rightarrow \frac{1}{tan\theta }\left(\frac{2tan\theta }{1-ta{n}^2\theta }\right)(1+\frac{1}{cos{2}^2\theta })(1+sec8\theta )...(1+sec{2}^n\theta )$

$\Rightarrow cot\theta .tan2\theta (1+\frac{1+ta{n}^{2}2\theta }{1-ta{n}^{2}2\theta })(1+sec8\theta )...(1+sec{2}^n\theta )$

$\Rightarrow cot\theta \left(\frac{2tan2\theta }{1-ta{n}^{2}2\theta }\right)(1+\frac{1}{cos{2}^3\theta })...(1+sec{2}^n\theta )$

$\Rightarrow cot\theta tan4\theta .(1+\frac{1+ta{n}^{2}4\theta }{1-ta{n}^{2}4\theta })...(1+sec{2}^2\theta )$

If we go n times ellipse this

$\Rightarrow cot\theta .tan{2}^n\theta .$

$\Rightarrow \frac{tan{2}^n\theta }{tan{2}^0\theta }$ R.H.S