Question

Prove that $ \dfrac{1}{{1 + \sin \theta }} + \dfrac{1}{{1 - \sin \theta }} =2{\sec ^2}\theta$

Answer 1

Hint: Here the given question is to prove the given expression, here we need to solve the expression by solving fraction and addition, then by suitable replacement we can reach the desired solution, here we know that fraction addition is solved by solving denominators collectively.
Formulae Used:
\[\Rightarrow \left(a+b\right) \left(a-b\right)=\left(a^2-b^2\right)\]
\[\Rightarrow \dfrac{1}{{1 + \sin \theta }} + \dfrac{1}{{1 - \sin \theta }} = 2{\sec ^2}\theta \]
\[ \Rightarrow {\sin ^2}\theta + {\cos ^2}\theta = 1\]
\[ \Rightarrow \sec \theta = \dfrac{1}{{\cos \theta }}\]

Complete step-by-step solution:
Here in the above question we need to solve the fraction, and then simplify for the required proof as demand in the question, on solving we get:
Here to solve left hand side of the question, we are going to solve the denominators, in order to make the denominator same of both the fractions and then solve with numerator in order to solve the question, on solving we get:
\[
   \Rightarrow \dfrac{1}{{1 + \sin \theta }} + \dfrac{1}{{1 - \sin \theta }} \\
= \dfrac{{\left( {1 - \sin \theta } \right) + \left( {1 + \sin \theta } \right)}}{{\left( {1 + \sin \theta } \right)\left( {1 - \sin \theta } \right)}} \\
= \dfrac{{1 - \sin \theta + 1 + \sin \theta }}{{{1^2} - {{\sin }^2}\theta }}\,\left( {u\sin g\,(a + b)(a - b) = {a^2} - {b^2}} \right) \\
= \dfrac{{1 + 1}}{{1 - {{\sin }^2}\theta }} = \dfrac{2}{{{{\cos }^2}\theta }}\,\left( {u\sin g\,{{\sin }^2}\theta + {{\cos }^2}\theta = 1} \right) \\
  =2{\sec ^2}\theta \,(\sec \theta = \dfrac{1}{{\cos \theta }}) \\
 \]
Here we can see that the left side of the equation is equal to the right hand side of the equation, hence proved.

Note: Here in the given question, we need to use trigonometric formulae, because in such a question in which we need to prove, we need to do a replacement and accordingly we need to use the required formulae to solve.