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,
\[\begin{array}{l}
 = 1 + \dfrac{{2{{\tan }^2}\theta }}{{1 - {{\tan }^2}\theta }}\\
 = \dfrac{{1 - {{\tan }^2}\theta + 2{{\tan }^2}\theta }}{{1 - {{\tan }^2}\theta }}\\
 = \dfrac{{1 + {{\tan }^2}\theta }}{{1 - {{\tan }^2}\theta }}\\
Now,\tan \theta = \dfrac{{\sin \theta }}{{\cos \theta }}\\
\therefore \dfrac{{1 + \left( {\dfrac{{{{\sin }^2}\theta }}{{{{\cos }^2}\theta }}} \right)}}{{1 - \left( {\dfrac{{{{\sin }^2}\theta }}{{{{\cos }^2}\theta }}} \right)}}\\
 = \dfrac{{\dfrac{{{{\cos }^2}\theta + {{\sin }^2}\theta }}{{{{\cos }^2}\theta }}}}{{\dfrac{{{{\cos }^2}\theta - {{\sin }^2}\theta }}{{{{\cos }^2}\theta }}}}\\
 = \dfrac{{{{\cos }^2}\theta + {{\sin }^2}\theta }}{{{{\cos }^2}\theta - {{\sin }^2}\theta }}
\end{array}\]
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.