Question

Prove that $\left( {1 + \sec 2\theta } \right)\left( {1 + \sec 4\theta } \right)\left( {1 + \sec 8\theta } \right)......\left( {1 + \sec {2^n}\theta } \right) = \tan \left( {{2^n}\theta } \right) \cdot \cot \theta$

Answer 1

To prove the given trigonometric functions, consider the LHS terms and apply the trigonometric identities functions in which as we know $\sec 2\theta = \dfrac{1}{{\cos 2\theta }}$, in same way we can apply for $\sec 4\theta$ and $\sec 8\theta$, hence further simplifying the terms we get LHS = RHS.

Complete step by step solution:
$\left( {1 + \sec 2\theta } \right)\left( {1 + \sec 4\theta } \right)\left( {1 + \sec 8\theta } \right)......\left( {1 + \sec {2^n}\theta } \right) = \tan \left( {{2^n}\theta } \right) \cdot \cot \theta$
Let us consider the LHS terms as:
$\left( {1 + \sec 2\theta } \right)\left( {1 + \sec 4\theta } \right)\left( {1 + \sec 8\theta } \right)......\left( {1 + \sec {2^n}\theta } \right)$
Applying the trigonometric identities to the terms we, we know that $\sec 2\theta = \dfrac{1}{{\cos 2\theta }}$, hence the equation is:
$\left( {1 + \dfrac{1}{{\cos 2\theta }}} \right)\left( {1 + \sec 4\theta } \right)\left( {1 + \sec 8\theta } \right)......\left( {1 + \sec {2^n}\theta } \right)$
$\Rightarrow\left( {1 + \dfrac{{1 + {{\tan }^2}\theta }}{{1 - {{\tan }^2}\theta }}} \right)\left( {1 + \sec 4\theta } \right)\left( {1 + \sec 8\theta } \right)......\left( {1 + \sec {2^n}\theta } \right)$
Simplifying the trigonometric functions as:
$\left( {\dfrac{2}{{1 - {{\tan }^2}\theta }}} \right)\left( {1 + \sec 4\theta } \right)\left( {1 + \sec 8\theta } \right)......\left( {1 + \sec {2^n}\theta } \right)$
Now multiply and divide the terms with $\tan \theta$
$\dfrac{1}{{\tan \theta }}\left( {\dfrac{{2\tan \theta }}{{1 - {{\tan }^2}\theta }}} \right)\left( {1 + \dfrac{1}{{\cos {2^2}\theta }}} \right)\left( {1 + \sec 8\theta } \right)......\left( {1 + \sec {2^n}\theta } \right)$
$\Rightarrow\cot \theta \cdot \tan 2\theta \left( {1 + \dfrac{{1 + {{\tan }^2}2\theta }}{{1 - {{\tan }^2}2\theta }}} \right)\left( {1 + \sec 8\theta } \right)......\left( {1 + \sec {2^n}\theta } \right)$
$\Rightarrow\cot \theta \left( {\dfrac{{2\tan 2\theta }}{{1 - {{\tan }^2}2\theta }}} \right)\left( {1 + \dfrac{1}{{\cos {2^3}\theta }}} \right)......\left( {1 + \sec {2^n}\theta } \right)$
$\Rightarrow\cot \theta \cdot \tan 4\theta \left( {1 + \dfrac{{1 + {{\tan }^2}4\theta }}{{1 - {{\tan }^2}4\theta }}} \right)......\left( {1 + \sec {2^2}\theta } \right)$
If we go n times ellipse then
$\cot \theta \cdot \tan {2^n}\theta$
$\therefore$ $\dfrac{{\tan {2^n}\theta }}{{\tan {2^0}\theta }}$ = RHS

As, LHS = RHS.Hence it is proved.

Note: The key point to prove trigonometric functions is that we must know all the trigonometric identity functions, to solve the terms in the given expression such that to prove LHS = RHS we must consider any of the terms of side and must know all the basic identities and relation between the trigonometric functions.