Question

\[\frac{\tan^2 A}{(\sec A + 1)^2}\] is equal to ______.

Options

• \[\frac{1 + \cos A}{1 - \cos A}\]

• \[\frac{1 - \cos A}{1 + \cos A}\]

• \[\frac{1}{1 + \cos A}\]

• \[\frac{1}{1 - \cos A}\]

Answer 1

\[\frac{\tan^2 A}{(\sec A + 1)^2}\] is equal to \[\underline{\boldsymbol{\frac{1 - \cos A}{1 + \cos A}}}\].

Explanation:

Solving,

\[\Rightarrow \frac{\tan^2 A}{(\sec A + 1)^2}\]

\[\Rightarrow \frac{\sec^2 A - 1}{(\sec A + 1)^2}\]

\[\Rightarrow \frac{(\sec A + 1)(\sec A - 1)}{(\sec A + 1)^2}\]

\[\Rightarrow \frac{\sec A - 1}{\sec A + 1}\]

\[\Rightarrow \frac{\frac{1}{\cos A} - 1}{\frac{1}{\cos A} + 1}\]

\[\Rightarrow \frac{\frac{1 - \cos A}{\cos A}}{\frac{1 + \cos A}{\cos A}}\]

\[\Rightarrow \frac{1 - \cos A}{1 + \cos A}\]