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Prove that $\sqrt {\dfrac{{1 + \sin A}}{{1 - \sin A}}} = \sec A + \tan A$

Answer Verified Verified
Hint: Rationalize the given equation and simplify.

 Consider $L.H.S = \sqrt {\dfrac{{1 + \sin A}}{{1 - \sin A}}} $
Now, rationalize the denominator i.e. in square root multiply and divide by $\left( {1 + \sin A} \right)$
$ \Rightarrow \sqrt {\dfrac{{1 + \sin A}}{{1 - \sin A}}} = \sqrt {\dfrac{{\left( {1 + \sin A} \right) \times \left( {1 + \sin A} \right)}}{{\left( {1 - \sin A} \right) \times \left( {1 + \sin A} \right)}}} $
In denominator it is $\left( {{a^2} - {b^2}} \right)$form
$ \Rightarrow \sqrt {\dfrac{{\left( {1 + \sin A} \right) \times \left( {1 + \sin A} \right)}}{{\left( {1 - \sin A} \right) \times \left( {1 + \sin A} \right)}}} = \sqrt {\dfrac{{{{\left( {1 + \sin A} \right)}^2}}}{{{1^2} - {{\sin }^2}A}}} $
Now we know$1 - {\sin ^2}A = {\cos ^2}A$so, substitute this value
\[
   \Rightarrow \sqrt {\dfrac{{{{\left( {1 + \sin A} \right)}^2}}}{{{1^2} - {{\sin }^2}A}}} = \sqrt {\dfrac{{{{\left( {1 + \sin A} \right)}^2}}}{{{{\cos }^2}A}}} = \dfrac{{1 + \sin A}}{{\cos A}} \\
   \Rightarrow \dfrac{1}{{\cos A}} + \dfrac{{\sin A}}{{\cos A}}.................\left( 1 \right) \\
\]
Now, we know $\dfrac{1}{{\cos A}} = \sec A,{\text{ }}\dfrac{{\sin A}}{{\cos A}} = \tan A$
From equation 1
\[
   \Rightarrow \dfrac{1}{{\cos A}} + \dfrac{{\sin A}}{{\cos A}} = \sec A + \tan A \\
  {\text{ }} = R.H.S \\
\]
Hence Proved.

Note: - In this type of question the key concept we have to remember is that always rationalize the denominator, then simplify using general trigonometric formulas we will get the required answer.

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