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

Hint: 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.