Question

Find the value of $\theta $, if
$\dfrac{{\cos \theta }}{{1 - \sin \theta }} + \dfrac{{\cos \theta }}{{1 + \sin \theta }} = 4$ , $\theta \leqslant {90^ \circ }$

Answer 1

Hint: First we will take LCM for the simplification of the equation then using formulas like ${\cos ^2}\theta = 1 - {\sin ^2}\theta $ we will get an equation in $\theta $ or in the function of $\theta $.
Solving that equation we will get the values of $\theta $ hence will obtain the solutions of the given equation.

Complete step-by-step answer:
Given data: $\dfrac{{\cos \theta }}{{1 - \sin \theta }} + \dfrac{{\cos \theta }}{{1 + \sin \theta }} = 4$
Solving for the given equation, we get
i.e. $\dfrac{{\cos \theta }}{{1 - \sin \theta }} + \dfrac{{\cos \theta }}{{1 + \sin \theta }} = 4$
Taking LCM we get,
$ \Rightarrow \dfrac{{\cos \theta \left( {1 + \sin \theta } \right)}}{{\left( {1 - \sin \theta } \right)\left( {1 + \sin \theta } \right)}} + \dfrac{{\cos \theta \left( {1 - \sin \theta } \right)}}{{\left( {1 - \sin \theta } \right)\left( {1 + \sin \theta } \right)}} = 4$
Simplifying the numerator as the denominator are equal, we get
$ \Rightarrow \dfrac{{\cos \theta \left( {1 + \sin \theta } \right) + \cos \theta \left( {1 - \sin \theta } \right)}}{{\left( {1 - \sin \theta } \right)\left( {1 + \sin \theta } \right)}} = 4$
Simplifying the brackets we get,
$ \Rightarrow \dfrac{{\cos \theta + \sin \theta \cos \theta + \cos \theta - \sin \theta \cos \theta }}{{\left( {1 - \sin \theta } \right)\left( {1 + \sin \theta } \right)}} = 4$
Simplifying the bracket terms in the denominator, we get
$ \Rightarrow \dfrac{{2\cos \theta }}{{1 - {{\sin }^2}\theta }} = 4$
We know that ${\cos ^2}\theta = 1 - {\sin ^2}\theta $therefore using this we get,
$ \Rightarrow \dfrac{{2\cos \theta }}{{{{\cos }^2}\theta }} = 4.................(i)$
Multiplying both sides by \[{\cos ^2}\theta \], we get
$ \Rightarrow 2\cos \theta = 4{\cos ^2}\theta $
Dividing both sides by 2, we get
$ \Rightarrow \cos \theta = 2{\cos ^2}\theta .................(ii)$
Taking the term on one side we get,
$ \Rightarrow 0 = 2{\cos ^2}\theta - \cos \theta $
Now, taking $\cos \theta $ common from both the terms we get,
$ \Rightarrow 0 = \cos \theta \left( {2\cos \theta - 1} \right)$
i.e. $2\cos \theta - 1 = 0$ or $\cos \theta = 0$
$\therefore \cos \theta = \dfrac{1}{2}$ or $\cos \theta = 0$
Substituting $\dfrac{1}{2} = \cos {60^ \circ }$and $0 = \cos {90^ \circ }$, we get

$ \Rightarrow \cos \theta = \cos {60^ \circ }$
On comparing we get,
$ \Rightarrow \theta = {60^ \circ }$
And $\cos \theta = \cos {90^ \circ }$
On comparing we get,
$ \Rightarrow \theta = {90^ \circ }$
Therefore the solutions of the given equation will \[{60^ \circ }\,and\,\,{90^ \circ }\].

Note: Most of the students will cancel out $\cos \theta $in the equation(i) or equation(ii) but it will reduce a solution a from the final answer when we cancel out variable terms like this, a solution for the equation reduces as if cancel out $\cos \theta $, we would have reduced a solution i.e. $\cos \theta = 0$, so remember this point always not only for this question but for all the questions in which we have to find the solutions of a given equation as doing this number of real solution reduces.