QUESTION

# Prove that the value of $\sqrt {\dfrac{{1 + \sin \theta }}{{1 - \sin \theta }}} + \sqrt {\dfrac{{1 - \sin \theta }}{{1 + \sin \theta }}} = 2\sec \theta$

Hint: To solve these types of problems, let us rationalise the denominator and solve.

The LHS given is $\sqrt {\dfrac{{1 + \sin \theta }}{{1 - \sin \theta }}} + \sqrt {\dfrac{{1 - \sin \theta }}{{1 + \sin \theta }}}$

So, let us now rationalise the denominators in both the terms
So, we get $\sqrt {\dfrac{{1 - \sin \theta }}{{1 + \sin \theta }} \times \dfrac{{1 - \sin \theta }}{{1 - \sin \theta }}} + \sqrt {\dfrac{{1 + \sin \theta }}{{1 - \sin \theta }} \times \dfrac{{1 + \sin \theta }}{{1 + \sin \theta }}}$
$= \sqrt {\dfrac{{{{\left( {1 - \sin \theta } \right)}^2}}}{{1 - {{\sin }^2}\theta }}} + \sqrt {\dfrac{{{{\left( {1 + \sin \theta } \right)}^2}}}{{1 - {{\sin }^2}\theta }}} \\ \\$
We know that ${\sin ^2}\theta + {\cos ^2}\theta = 1 \\ \\$
$\Rightarrow 1 - {\sin ^2}\theta = {\cos ^2}\theta \\ \Rightarrow \sqrt {1 - {{\sin }^2}\theta } = \cos \theta \\$
So, we get the equation as
$\dfrac{{1 - \sin \theta }}{{\cos \theta }} + \dfrac{{1 + \sin \theta }}{{\cos \theta }}$
On adding the two terms, we get
$\dfrac{{1 - \sin \theta + 1 + \sin \theta }}{{\cos \theta }} \\ = \dfrac{2}{{\cos \theta }} \\$
Since, cos and sec are reciprocal of each other, we can write the equation as
$2\sec \theta$ =RHS
Hence , the result is proved.

Note: Whenever, solving these type of problems, make use of the appropriate trigonometric identities in accordance to the RHS which has to be proved.