Question

Find the value of : $3\tan \theta + \cot \theta = 5\cos ec\theta$

Answer 1

According to the question, convert the given equation into $sin\theta$ and $\cos \theta$. After converting, solve the required equation using trigonometric formulas to get the result.

Formula used:
Here, formula to be used is ${\sin ^2}\theta = 1 - {\cos ^2}\theta$and conversion of $\tan \theta$ , $\cot \theta$ and $\cos ec\theta$ will be used.

Complete step-by-step answer:
Firstly, convert both right hand side and left hand side in terms of $sin\theta$ and $\cos \theta$. As, we know $\tan \theta = \dfrac{{\sin \theta }}{{\cos \theta }}$ , $\cot \theta = \dfrac{{\cos \theta }}{{\sin \theta }}$ and $\cos ec\theta = \dfrac{1}{{\sin \theta }}$ .
Given, $3\tan \theta + \cot \theta = 5\cos ec\theta$
Put all the values of $\tan \theta$ , $\cot \theta$ and $\cos ec\theta$ in the above equation.
We get, $3\left( {\dfrac{{\sin \theta }}{{\cos \theta }}} \right) + \dfrac{{\cos \theta }}{{\sin \theta }} = 5\left( {\dfrac{1}{{\sin \theta }}} \right)$
Now, we will open all the brackets
$\Rightarrow$ $\dfrac{{3\sin \theta }}{{\cos \theta }} + \dfrac{{\cos \theta }}{{\sin \theta }} = \dfrac{5}{{sin\theta }}$
Here, we will take the L.C.M on the left side of the above equation.
Therefore, we get
$\Rightarrow$ $\dfrac{{{\mathop{\rm Sin}\nolimits} \theta \left( {3\sin \theta } \right) + \left( {\cos \theta } \right)\cos \theta }}{{\cos \theta \sin \theta }} = \dfrac{5}{{sin\theta }}$
On simplifying,
$\Rightarrow$ $\dfrac{{3{{{\mathop{\rm Sin}\nolimits} }^2}\theta + {{\cos }^2}\theta }}{{\cos \theta \sin \theta }} = \dfrac{5}{{sin\theta }}$
Again, on cancelling $\sin \theta$ from the denominator from both sides.
We get, $\dfrac{{3{{{\mathop{\rm Sin}\nolimits} }^2}\theta + {{\cos }^2}\theta }}{{\cos \theta }} = 5$
By cross multiplication:
$\Rightarrow$ $3{{\mathop{\rm Sin}\nolimits} ^2}\theta + {\cos ^2}\theta = 5\cos \theta$
Now, taking all the terms on the left hand side to make the right hand 0.
$\Rightarrow$ $3{{\mathop{\rm Sin}\nolimits} ^2}\theta + {\cos ^2}\theta - 5\cos \theta = 0$
Converting all the terms in terms of $\cos \theta$ by using the identity ${\sin ^2}\theta = 1 - {\cos ^2}\theta$ which comes from the identity ${\sin ^2}\theta + {\cos ^2}\theta = 1$.
So, by substituting the value of ${\sin ^2}\theta$ we get,
$3\left( {1 - {{\cos }^2}\theta } \right) + {\cos ^2}\theta - 5\cos \theta = 0$
$\Rightarrow$ $3 - 3{\cos ^2}\theta + {\cos ^2}\theta - 5\cos \theta = 0$
On further simplifying we get,
$\Rightarrow$ $3 - 2{\cos ^2}\theta - 5\cos \theta = 0$ which can be written as $2{\cos ^2}\theta + 5\cos \theta - 3 = 0$
By using factorisation method we can this quadratic equation:
$\Rightarrow$ $2{\cos ^2}\theta + 6\cos \theta - \cos \theta - 3 = 0$
Taking out all the common values,
We get, $2\cos \theta \left( {\cos \theta + 3} \right) - 1\left( {\cos \theta + 3} \right) = 0$
Therefore, $\left( {\cos \theta + 3} \right)\left( {2\cos \theta - 1} \right) = 0$
Putting both the values separately equal to 0.
Firstly take, $\left( {\cos \theta + 3} \right) = 0$
Here, $\cos \theta = - 3$ which is not possible.
Secondly take, $\left( {2\cos \theta - 1} \right) = 0$
Here, $\cos \theta = \dfrac{1}{2}$
As at $\cos \theta = \dfrac{1}{2}$ , $\theta$ comes out to be 60 which is $\dfrac{\pi }{3}$
Or we can write in this form also, $\theta = 2n\pi \pm \dfrac{\pi }{3}$

Note: In these types of questions, we need to check whether the equation is in the form of sin and cos, as it makes the question simpler to solve. As well as we have all the trigonometric formulas in this form. So, the required value can be easily calculated.