Question

If $\sqrt 3 \sin \theta = \cos \theta$ , find the value of $\dfrac{{\sin \theta \tan \theta \left( {1 + \cos \theta } \right)}}{{\sin \theta + \cos \theta }}$ .

Answer 1

Firstly, find the values of $\sin \theta$ and $\cos \theta$ , using the relation $\sqrt 3 \sin \theta = \cos \theta$ .
Then, substitute the values of $\sin \theta ,\cos \theta ,\tan \theta$ in $\dfrac{{\sin \theta \tan \theta \left( {1 + \cos \theta } \right)}}{{\sin \theta + \cos \theta }}$ , to get the required answer.

Complete step-by-step answer:
It is given that, $\sqrt 3 \sin \theta = \cos \theta$ .

$$\therefore \dfrac{{\sin \theta }}{{\cos \theta }} = \dfrac{1}{{\sqrt 3 }} \\\ \Rightarrow \tan \theta = \dfrac{1}{{\sqrt 3 }} \\\ \Rightarrow \theta = {\tan ^{ - 1}}\dfrac{1}{{\sqrt 3 }} \\\ \Rightarrow \theta = 30^\circ \\\$$

Now, we need to find the value of $\dfrac{{\sin \theta \tan \theta \left( {1 + \cos \theta } \right)}}{{\sin \theta + \cos \theta }}$ . So, firstly we need to find the values of $\sin \theta$ and $\cos \theta$ by putting $\theta = 30^\circ$ .
$\Rightarrow \sin \theta = \sin 30^\circ = \dfrac{1}{2}$ and $\cos \theta = \cos 30^\circ = \dfrac{{\sqrt 3 }}{2}$ .
Thus, we get $\tan \theta = \dfrac{1}{{\sqrt 3 }},\sin \theta = \dfrac{1}{2}$ and $\cos \theta = \dfrac{{\sqrt 3 }}{2}$ .
Now, we will substitute the values $\tan \theta = \dfrac{1}{{\sqrt 3 }},\sin \theta = \dfrac{1}{2}$ and $\cos \theta = \dfrac{{\sqrt 3 }}{2}$ in $\dfrac{{\sin \theta \tan \theta \left( {1 + \cos \theta } \right)}}{{\sin \theta + \cos \theta }}$ , to get the required answer.
$\Rightarrow \dfrac{{\sin \theta \tan \theta \left( {1 + \cos \theta } \right)}}{{\sin \theta + \cos \theta }} = \dfrac{{\left( {\dfrac{1}{2}} \right)\left( {\dfrac{1}{{\sqrt 3 }}} \right)\left( {1 + \dfrac{{\sqrt 3 }}{2}} \right)}}{{\dfrac{1}{2} + \dfrac{{\sqrt 3 }}{2}}} \\\ = \dfrac{{\dfrac{1}{{2\sqrt 3 }}\left( {\dfrac{{2 + \sqrt 3 }}{2}} \right)}}{{\dfrac{{1 + \sqrt 3 }}{2}}} \\\ = \dfrac{1}{{2\sqrt 3 }} \times \dfrac{{2 + \sqrt 3 }}{{1 + \sqrt 3 }} \\\ = \dfrac{{2 + \sqrt 3 }}{{2\sqrt 3 \left( {1 + \sqrt 3 } \right)}} \\\$
Thus, we get the value $\dfrac{{\sin \theta \tan \theta \left( {1 + \cos \theta } \right)}}{{\sin \theta + \cos \theta }} = \dfrac{{2 + \sqrt 3 }}{{2\sqrt 3 \left( {1 + \sqrt 3 } \right)}}$ .

Note: Important table to be remembered:

| $0^\circ$ | $30^\circ$| $45^\circ$| $60^\circ$| $90^\circ$
---|---|---|---|---|---
$\sin \theta$| 0| $\dfrac{1}{2}$ | $\dfrac{1}{{\sqrt 2 }}$ | $\dfrac{{\sqrt 3 }}{2}$| 1
$\operatorname{cosec} \theta$| Indeterminate| 2| $\sqrt 2$ | $\dfrac{2}{{\sqrt 3 }}$| 1
$\cos \theta$| 1| $\dfrac{{\sqrt 3 }}{2}$ | $\dfrac{1}{{\sqrt 2 }}$| $\dfrac{1}{2}$| 0
$\sec \theta$| 1| $\dfrac{2}{{\sqrt 3 }}$ | $\sqrt 2$| 2| Indeterminate
$\tan \theta$| 0| $\dfrac{1}{{\sqrt 3 }}$ | 1| $\sqrt 3$| Indeterminate
$\cot \theta$| Indeterminate| $\sqrt 3$ | 1| $\dfrac{1}{{\sqrt 3 }}$| 0