Question

Prove the following: $1 + \tan 2\theta \tan \theta = \sec 2\theta$

Answer 1

Hint: We know that $\tan \theta = \dfrac{{\sin \theta }}{{\cos \theta }}\& \tan 2\theta = \dfrac{{2\tan \theta }}{{1 - {{\tan }^2}\theta }}$ Use these formulas wherever necessary to prove the required solution. And also some other trigonometric identities will also be needed to complete the problem.
Complete step by step answer:
First of all break $\tan 2\theta = \dfrac{{2\tan \theta }}{{1 - {{\tan }^2}\theta }}$
Now LHS becomes,

= 1 + \dfrac{{2{{\tan }^2}\theta }}{{1 - {{\tan }^2}\theta }}\\\ = \dfrac{{1 - {{\tan }^2}\theta + 2{{\tan }^2}\theta }}{{1 - {{\tan }^2}\theta }}\\\ = \dfrac{{1 + {{\tan }^2}\theta }}{{1 - {{\tan }^2}\theta }}\\\ Now,\tan \theta = \dfrac{{\sin \theta }}{{\cos \theta }}\\\ \therefore \dfrac{{1 + \left( {\dfrac{{{{\sin }^2}\theta }}{{{{\cos }^2}\theta }}} \right)}}{{1 - \left( {\dfrac{{{{\sin }^2}\theta }}{{{{\cos }^2}\theta }}} \right)}}\\\ = \dfrac{{\dfrac{{{{\cos }^2}\theta + {{\sin }^2}\theta }}{{{{\cos }^2}\theta }}}}{{\dfrac{{{{\cos }^2}\theta - {{\sin }^2}\theta }}{{{{\cos }^2}\theta }}}}\\\ = \dfrac{{{{\cos }^2}\theta + {{\sin }^2}\theta }}{{{{\cos }^2}\theta - {{\sin }^2}\theta }} \end{array}$$ Now, we know that $${\cos ^2}\theta + {\sin ^2}\theta = 1\& {\cos ^2}\theta - {\sin ^2}\theta = \cos 2\theta $$ Using these trigonometric identities we can rewrite the last step as $$\begin{array}{l} = \dfrac{1}{{\cos 2\theta }}\\\ = \sec 2\theta \end{array}$$ Hence LHS=RHS (proved) Note: Do note that $$\dfrac{1}{{\cos 2\theta }} = \sec 2\theta $$ because secant is the reciprocal of cosine and also cosecant is the reciprocal of sine. We can use the formula of $$\cos 2\theta = \dfrac{{1 - {{\tan }^2}\theta }}{{1 + {{\tan }^2}\theta }}$$ direct to minimize calculation. Use the properties of trigonometry wisely to calculate this minimalistically.