Question

Prove the following statement:
\[\sqrt{\dfrac{1-\sin A}{1+\sin A}}=\sec A-\tan A\]

Answer 1

Hint: First of all consider the LHS of the given equation and multiply it by \[\sqrt{\dfrac{1-\sin A}{1-\sin A}}\]. Now, use \[1-{{\sin }^{2}}\theta ={{\cos }^{2}}\theta \] and simplify the given expression. Now, separate the terms to prove the desired result.
Complete step-by-step answer:
Here, we have to prove that \[\sqrt{\dfrac{1-\sin A}{1+\sin A}}=\sec A-\tan A\]. Let us consider the LHS of the equation given in the question.
\[E=\sqrt{\dfrac{1-\sin A}{1+\sin A}}\]
By multiplying the above expression by \[\sqrt{\dfrac{1-\sin A}{1-\sin A}}\], we get,
\[E=\sqrt{\dfrac{1-\sin A}{1+\sin A}}.\sqrt{\dfrac{1-\sin A}{1-\sin A}}\]
We know that \[\sqrt{a}.\sqrt{b}=\sqrt{ab}\]. By using this in the above expression, we get,
\[E=\sqrt{\dfrac{\left( 1-\sin A \right)}{\left( 1+\sin A \right)}.\dfrac{\left( 1-\sin A \right)}{\left( 1-\sin A \right)}}\]
\[E=\sqrt{\dfrac{{{\left( 1-\sin A \right)}^{2}}}{\left( 1+\sin A \right)\left( 1-\sin A \right)}}\]
We know that \[\left( a+b \right)\left( a-b \right)={{a}^{2}}-{{b}^{2}}\]. By using this in the above expression, we get,
\[E=\sqrt{\dfrac{{{\left( 1-\sin A \right)}^{2}}}{{{\left( 1 \right)}^{2}}-{{\left( \sin A \right)}^{2}}}}\]
\[E=\sqrt{\dfrac{{{\left( 1-\sin A \right)}^{2}}}{1-{{\sin }^{2}}A}}\]
We know that \[{{\cos }^{2}}\theta +{{\sin }^{2}}\theta =1\] or \[1-{{\sin }^{2}}\theta ={{\cos }^{2}}\theta \]. By using this in the above expression, we get,
\[E=\sqrt{\dfrac{{{\left( 1-\sin A \right)}^{2}}}{{{\cos }^{2}}A}}\]
We know that \[\left( \dfrac{{{a}^{2}}}{{{b}^{2}}} \right)={{\left( \dfrac{a}{b} \right)}^{2}}\]. By using this in the above expression, we get,
\[E=\sqrt{{{\left( \dfrac{1-\sin A}{\cos A} \right)}^{2}}}\]
We know that \[\sqrt{{{x}^{2}}}=x\]. By using this in the above expression, we get,
\[E=\dfrac{1-\sin A}{\cos A}\]
We can also write the above expression as,
\[E=\dfrac{1}{\cos A}-\dfrac{\sin A}{\cos A}\]
We know that \[\dfrac{1}{\cos \theta }=\sec \theta \] and \[\dfrac{\sin \theta }{\cos \theta }=\tan \theta \]. By using this in the above expression, we get,
E = sec A – tan A
E = RHS
So, we get, LHS = RHS
Hence, we have proved that \[\sqrt{\dfrac{1-\sin A}{1+\sin A}}=\sec A-\tan A\].

Note: In these type of questions containing the square root, some students try to convert the given angles into half angles but that is reliable in case of \[\cos \theta \] but in case of \[\sin \theta \] we should try to remove the square root by multiplying the conjugate of the denominator that is if the denominator is a + b, we will multiply (a – b) and vice versa. Then use trigonometric formulas to further simplify it.