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

Hint: 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