Prove the following
\[\left( 1+\sec 2\theta \right)\left( 1+\sec 4\theta \right).....\left( 1+\sec {{2}^{n}}\theta \right)=\tan {{2}^{n}}\theta \cot \theta \]

Hint: First of all, consider the LHS of the given equation. Now, use \[\sec \theta =\dfrac{1}{\cos \theta }\] and \[\cos 2\theta =\dfrac{1-{{\tan }^{2}}\theta }{1+{{\tan }^{2}}\theta }.\] Now, multiply and divide by \[\tan \theta \] and use \[\tan 2\theta =\dfrac{2\tan \theta }{1-{{\tan }^{2}}\theta }.\] Then repeat these steps for each term and prove the given result.

Complete step-by-step solution:
In this question, we have to prove that \[\left( 1+\sec 2\theta \right)\left( 1+\sec 4\theta \right).....\left( 1+\sec {{2}^{n}}\theta \right)=\tan {{2}^{n}}\theta \cot \theta .\] Let us consider the LHS of the given equation.
\[LHS=\left( 1+\sec 2\theta \right)\left( 1+\sec 4\theta \right)\left( 1+\sec 8\theta \right).....\left( 1+\sec {{2}^{n}}\theta \right)\]
We know that \[\sec \theta =\dfrac{1}{\cos \theta }\] so by using this, we get,
\[\Rightarrow LHS=\left( 1+\dfrac{1}{\cos 2\theta } \right)\left( 1+\sec 4\theta \right)\left( 1+\sec 8\theta \right).....\left( 1+\sec {{2}^{n}}\theta \right)\]
We know that \[\cos 2\theta =\dfrac{1-{{\tan }^{2}}\theta }{1+{{\tan }^{2}}\theta }.\] So, by using this, we get,
\[\Rightarrow LHS=\left( 1+\dfrac{1+{{\tan }^{2}}\theta }{1-{{\tan }^{2}}\theta } \right)\left( 1+\sec 4\theta \right)\left( 1+\sec 8\theta \right).....\left( 1+\sec {{2}^{n}}\theta \right)\]
By simplifying the above expression, we get,
\[\Rightarrow LHS=\left( \dfrac{2}{1-{{\tan }^{2}}\theta } \right)\left( 1+\sec 4\theta \right)\left( 1+\sec 8\theta \right).....\left( 1+\sec {{2}^{n}}\theta \right)\]
By multiplying and dividing \[\tan \theta \] in the above expression, we get,
\[\Rightarrow LHS=\dfrac{1}{\tan \theta }\left( \dfrac{2\tan \theta }{1-{{\tan }^{2}}\theta } \right)\left( 1+\sec 4\theta \right)\left( 1+\sec 8\theta \right).....\left( 1+\sec {{2}^{n}}\theta \right)\]
We know that \[\dfrac{1}{\tan \theta }=\cot \theta \] and \[\dfrac{2\tan \theta }{1-{{\tan }^{2}}\theta }=\tan 2\theta .\] By using these, we get,
\[\Rightarrow LHS=\left( \cot \theta \right)\left( \tan 2\theta \right)\left( 1+\sec 4\theta \right)\left( 1+\sec 8\theta \right).....\left( 1+\sec {{2}^{n}}\theta \right)\]
Again by repeating the above steps for \[\sec 4\theta ,\] we get,
\[\Rightarrow LHS=\left( \cot \theta \right)\left( \tan 2\theta \right)\left( 1+\dfrac{1+{{\tan }^{2}}2\theta }{1-{{\tan }^{2}}2\theta } \right)\left( 1+\sec 8\theta \right).....\left( 1+\sec {{2}^{n}}\theta \right)\]
\[\Rightarrow LHS=\left( \cot \theta \right)\left( \dfrac{2\tan 2\theta }{1-{{\tan }^{2}}2\theta } \right)\left( 1+\sec 8\theta \right).....\left( 1+\sec {{2}^{n}}\theta \right)\]
\[\Rightarrow LHS=\left( \cot \theta \right)\left( \tan 4\theta \right)\left( 1+\sec 8\theta \right).....\left( 1+\sec {{2}^{n}}\theta \right)\]
As we have got \[\cot \theta .\tan 2\theta \] for \[\left( 1+\sec 2\theta \right)\] and \[\cot \theta .\tan 4\theta \] for \[\left( 1+\sec 2\theta \right)\left( 1+\sec 4\theta \right)\] and \[\cot \theta .\tan 8\theta \] for \[\left( 1+{{\sec }^{2}}\theta \right)\left( 1+\sec 4\theta \right)\left( 1+\sec 8\theta \right).\] So, in the similar way, we get,
\[\Rightarrow LHS=\left( 1+\sec 2\theta \right)\left( 1+\sec 4\theta \right).....\left( 1+\sec {{2}^{n}}\theta \right)\]
\[\Rightarrow LHS=\cot \theta .\tan \left( {{2}^{n}}\theta \right)=RHS\]
Hence, we have proved that \[\left( 1+\sec 2\theta \right)\left( 1+\sec 4\theta \right).....\left( 1+\sec {{2}^{n}}\theta \right)=\tan {{2}^{n}}\theta \cot \theta .\]

Note: First of all, students must take care of the angles while using the double angle or half-angle formulas. For example, sometimes students make this mistake of writing \[\cos 4\theta =\dfrac{1-{{\tan }^{2}}\theta }{1+{{\tan }^{2}}\theta }\] while actually it is \[\cos 4\theta =\dfrac{1-{{\tan }^{2}}2\theta }{1+{{\tan }^{2}}2\theta }.\] So this must be taken care of. Also, it is advisable for students to memorize the formulas of \[\tan 2\theta ,\sin 2\theta ,\cos 2\theta \] in the terms of \[\sin \theta ,\cos \theta ,\tan \theta \] to easily solve these types of questions.