Question

Find the number of real roots of the equation $ \sec \theta +\operatorname{cosec}\theta =\sqrt{15} $ lying between 0 and $ \pi $ .

Answer 1

Hint: To find the number of real roots of the given equation first we will simplify the equation using trigonometric properties. Then we will solve the obtained quadratic equation by using the quadratic formula which is given as $ x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a} $ . Then compare the values with 0 and $ \pi $ to get the desired answer.

Complete step by step answer:
We have been given an equation $ \sec \theta +\operatorname{cosec}\theta =\sqrt{15} $ .
We have to find the number of real roots of the equation.
We have $ \sec \theta +\operatorname{cosec}\theta =\sqrt{15} $
We know that $ \sec \theta =\dfrac{1}{\cos \theta } $ and $ \operatorname{cosec}\theta =\dfrac{1}{\sin \theta } $
So, we get $ \dfrac{1}{\cos \theta }+\dfrac{1}{\sin \theta }=\sqrt{15} $
By taking LCM and solving further we get
 $ \begin{align}
  & \Rightarrow \dfrac{\sin \theta +\cos \theta }{\sin \theta \cos \theta }=\sqrt{15} \\
 & \Rightarrow \sin \theta +\cos \theta =\sqrt{15}\sin \theta \cos \theta \\
\end{align} $
Now, squaring both sides we get
 $ \Rightarrow {{\left( \sin \theta +\cos \theta \right)}^{2}}={{\left( \sqrt{15}\sin \theta \cos \theta \right)}^{2}} $
Now, we know that $ {{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab $
So, we get
\[\Rightarrow {{\sin }^{2}}\theta +{{\cos }^{2}}\theta +2\sin \theta \cos \theta ={{\left( \sqrt{15} \right)}^{2}}{{\left( \sin \theta \cos \theta \right)}^{2}}\]
Now, we know that $ {{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1 $
Now, by substituting the value we get
 $ \Rightarrow 1+2\sin \theta \cos \theta =15{{\left( \sin \theta \cos \theta \right)}^{2}} $
Now, we know that $ 2\sin \theta \cos \theta =\sin 2\theta $
Now, by multiplying the whole equation by 4 we get
 $ \Rightarrow 4\times 1+4\times 2\sin \theta \cos \theta =15{{\left( 2\times \sin \theta \cos \theta \right)}^{2}} $
Now, solving further we get
 $ \Rightarrow 4+4\sin 2\theta =15{{\left( \sin 2\theta \right)}^{2}} $
Now rearranging the equation we get
 $ \Rightarrow 15{{\left( \sin 2\theta \right)}^{2}}-4\sin 2\theta -4=0 $ which is a quadratic equation.
Now, solve the above equation by using the quadratic formula $ x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a} $
Now, putting the values in above formula we get
\[\begin{align}
  & \Rightarrow \sin 2\theta =\dfrac{-\left( -4 \right)\pm \sqrt{{{\left( -4 \right)}^{2}}-4\times 15\times \left( -4 \right)}}{2\times 15} \\
 & \Rightarrow \sin 2\theta =\dfrac{4\pm \sqrt{16+240}}{30} \\
 & \Rightarrow \sin 2\theta =\dfrac{4\pm \sqrt{256}}{30} \\
 & \Rightarrow \sin 2\theta =\dfrac{4\pm 16}{30} \\
\end{align}\]
Or we can write as
\[\Rightarrow \sin 2\theta =\dfrac{4+16}{30}\] and \[\Rightarrow \sin 2\theta =\dfrac{4-16}{30}\]
So, we get
\[\Rightarrow \sin 2\theta =\dfrac{20}{30}\] and \[\Rightarrow \sin 2\theta =\dfrac{-12}{30}\]
\[\Rightarrow \sin 2\theta =\dfrac{2}{3}\] and \[\Rightarrow \sin 2\theta =\dfrac{-2}{5}\]
So, the number of real roots of the equation $ \sec \theta +\operatorname{cosec}\theta =\sqrt{15} $ lying between 0 and $ \pi $ is one because negative value does not come between 0 and $ \pi $ .

Note:
To find the roots it is necessary to convert the given equation into quadratic form $ a{{x}^{2}}+bx+c=0 $ . A quadratic equation is also solved by using the factorization method. Also to solve such types of questions students must have knowledge of trigonometric properties.