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Question: Find the value of \(\theta \), if \(\dfrac{{\cos \theta }}{{1 - \sin \theta }} + \dfrac{{\cos \the...

Find the value of θ\theta , if
cos⁡θ1−sin⁡θ+cos⁡θ1+sin⁡θ=4\dfrac{{\cos \theta }}{{1 - \sin \theta }} + \dfrac{{\cos \theta }}{{1 + \sin \theta }} = 4 , θ⩽90∘\theta \leqslant {90^ \circ }

Explanation

Solution

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

Complete step-by-step answer:
Given data: cos⁡θ1−sin⁡θ+cos⁡θ1+sin⁡θ=4\dfrac{{\cos \theta }}{{1 - \sin \theta }} + \dfrac{{\cos \theta }}{{1 + \sin \theta }} = 4
Solving for the given equation, we get
i.e. cos⁡θ1−sin⁡θ+cos⁡θ1+sin⁡θ=4\dfrac{{\cos \theta }}{{1 - \sin \theta }} + \dfrac{{\cos \theta }}{{1 + \sin \theta }} = 4
Taking LCM we get,
⇒cos⁡θ(1+sin⁡θ)(1−sin⁡θ)(1+sin⁡θ)+cos⁡θ(1−sin⁡θ)(1−sin⁡θ)(1+sin⁡θ)=4\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
⇒cos⁡θ(1+sin⁡θ)+cos⁡θ(1−sin⁡θ)(1−sin⁡θ)(1+sin⁡θ)=4\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,
⇒cos⁡θ+sin⁡θcos⁡θ+cos⁡θ−sin⁡θcos⁡θ(1−sin⁡θ)(1+sin⁡θ)=4\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
⇒2cos⁡θ1−sin⁡2θ=4\Rightarrow \dfrac{{2\cos \theta }}{{1 - {{\sin }^2}\theta }} = 4
We know that cos⁡2θ=1−sin⁡2θ{\cos ^2}\theta = 1 - {\sin ^2}\theta therefore using this we get,
⇒2cos⁡θcos⁡2θ=4.................(i)\Rightarrow \dfrac{{2\cos \theta }}{{{{\cos }^2}\theta }} = 4.................(i)
Multiplying both sides by cos⁡2θ{\cos ^2}\theta , we get
⇒2cos⁡θ=4cos⁡2θ\Rightarrow 2\cos \theta = 4{\cos ^2}\theta
Dividing both sides by 2, we get
⇒cos⁡θ=2cos⁡2θ.................(ii)\Rightarrow \cos \theta = 2{\cos ^2}\theta .................(ii)
Taking the term on one side we get,
⇒0=2cos⁡2θ−cos⁡θ\Rightarrow 0 = 2{\cos ^2}\theta - \cos \theta
Now, taking cos⁡θ\cos \theta common from both the terms we get,
⇒0=cos⁡θ(2cos⁡θ−1)\Rightarrow 0 = \cos \theta \left( {2\cos \theta - 1} \right)
i.e. 2cos⁡θ−1=02\cos \theta - 1 = 0 or cos⁡θ=0\cos \theta = 0
∴cos⁡θ=12\therefore \cos \theta = \dfrac{1}{2} or cos⁡θ=0\cos \theta = 0
Substituting 12=cos⁡60∘\dfrac{1}{2} = \cos {60^ \circ }and 0=cos⁡90∘0 = \cos {90^ \circ }, we get

⇒cos⁡θ=cos⁡60∘ \Rightarrow \cos \theta = \cos {60^ \circ }
On comparing we get,
⇒θ=60∘\Rightarrow \theta = {60^ \circ }
And cos⁡θ=cos⁡90∘\cos \theta = \cos {90^ \circ }
On comparing we get,
⇒θ=90∘\Rightarrow \theta = {90^ \circ }
Therefore the solutions of the given equation will 60∘ and  90∘{60^ \circ }\,and\,\,{90^ \circ }.

Note: Most of the students will cancel out cos⁡θ\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⁡θ\cos \theta , we would have reduced a solution i.e. cos⁡θ=0\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.