Question

Show that $\sqrt {\dfrac{{1 + \sin \theta }}{{1 - \sin \theta }}} = \dfrac{{1 + \sin \theta }}{{\cos \theta }}$.

Answer 1

Hint:Multiply the numerator and denominator by $\sqrt {1 + \sin \theta }$ and use the identity $\left( {{{\sin }^2}\theta + {{\cos }^2}\theta } \right) = 1$ to get required answer.

Complete step-by-step answer:
Given that,
$\Rightarrow \sqrt {\dfrac{{1 + \sin \theta }}{{1 - \sin \theta }}}$
Multiplying the numerator and denominator by $\sqrt {1 + \sin \theta }$ we get:
$\Rightarrow \sqrt {\dfrac{{1 + \sin \theta }}{{1 - \sin \theta }}} \times \sqrt {\dfrac{{1 + \sin \theta }}{{1 + \sin \theta }}} \\\ \Rightarrow \sqrt {\dfrac{{{{\left( {1 + \sin \theta } \right)}^2}}}{{\left( {1 - {{\sin }^2}\theta } \right)}}} \\\$
As we know that:
$\left( {{{\sin }^2}\theta + {{\cos }^2}\theta } \right) = 1 \\\ \therefore 1 - {\sin ^2}\theta = {\cos ^2}\theta \\\$
After putting the value equation can be written as:
$\Rightarrow \sqrt {\dfrac{{{{\left( {1 + \sin \theta } \right)}^2}}}{{\left( {{{\cos }^2}\theta } \right)}}}$
Taking square root both sides we get:
$\dfrac{{1 + \sin \theta }}{{\cos \theta }}$
Hence proved that $\sqrt {\dfrac{{1 + \sin \theta }}{{1 - \sin \theta }}} = \dfrac{{1 + \sin \theta }}{{\cos \theta }}$

Note:- In this question after multiplying $\sqrt {\dfrac{{1 + \sin \theta }}{{1 - \sin \theta }}}$ by $\sqrt {\dfrac{{1 + \sin \theta }}{{1 + \sin \theta }}}$ we got $\sqrt {\dfrac{{{{\left( {1 + \sin \theta } \right)}^2}}}{{\left( {1 - {{\sin }^2}\theta } \right)}}}$ as we know that $\left( {a + b} \right)\left( {a - b} \right) = \left( {{a^2} - {b^2}} \right)$, after that we applied the trigonometric identity $\left( {{{\sin }^2}\theta + {{\cos }^2}\theta } \right) = 1$ then we took the square root both sides we got the result as $\dfrac{{1 + \sin \theta }}{{\cos \theta }}$, hence proved.Students should remember the important trigonometric identities and basic algebraic identities for solving these types of questions.