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

Question

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

Vedantu Content Team · Accepted Answer

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.

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

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

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