Question

Show that \[\sqrt{\dfrac{1+\sin A}{1-\sin A}}=\sec A+\tan A\].

Answer 1

Hint:
Firstly, rationalise the denominator of RHS, multiply with \[\sqrt{1+\sin A}\] in both numerator and denominator since in the numerator part there will be the square of rationalising factor which remove the radical sign in numerator and in denominator, we can multiply the terms over the square root and using identity we can write it as \[{{\cos }^{2}}A\] and the radical sign will be removed because of this square. After that separating the terms we will get the same as RHS and hence proved.

Complete step by step solution:
Taking LHS,
\[\sqrt{\dfrac{1+\sin A}{1-\sin A}}\]
Now rationalising the denominator part
\[\Rightarrow \sqrt{\dfrac{1+\sin A}{1-\sin A}}\times \dfrac{\sqrt{1+\sin A}}{\sqrt{1+\sin A}}\]
Now multiply within the square root
\[\Rightarrow \sqrt{\dfrac{{{(1+\sin A)}^{2}}}{(1-\sin A)(1+\sin A)}}\]
\[\Rightarrow \sqrt{\dfrac{{{(1+\sin A)}^{2}}}{(1-{{\sin }^{2}}A)}}\]
Now using the identity, \[{{\sin }^{2}}A+{{\cos }^{2}}A=1\] the denominator can be written as,
\[\Rightarrow \sqrt{\dfrac{{{(1+\sin A)}^{2}}}{{{\cos }^{2}}A}}\]
Also due to the square over the numerator and denominator, the radical sign will be removed
\[\Rightarrow \dfrac{1+\sin A}{\cos A}\]
Now on separating the above expression
\[\Rightarrow \dfrac{1}{\cos A}+\dfrac{\sin A}{\cos A}\]
Since we know that \[\dfrac{1}{\cos A}=\sec A\] and \[\dfrac{\sin A}{\cos A}=\tan A\]
Now replace that expression with above relation
\[\Rightarrow \sec A+\tan A\]
\[\Rightarrow \sec A+\tan A=RHS\]
Hence Proved

Note:
First of all, on seeing carefully we came to know that after rationalising the denominator of the LHS the modified denominator will be the same as the denominator of RHS means we can go ahead of this. Whenever we are given an expression of this type like square root in numerator and denominator with just opposite sign means on rationalising, we can simplify it then just do go with rationalising the denominator part.