Question & Answer
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 $

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

Complete step-by-step answer:

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.