Question

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

Answer 1

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.