Question

If $\tan \theta = - \dfrac{5}{{12}}$ , $\theta $ is not in the second quadrant, then show that $\dfrac{{\sin \left( {360^\circ - \theta } \right) + \tan \left( {90^\circ + \theta } \right)}}{{ - \sec \left( {270 + \theta } \right) + \operatorname{cosec} \left( { - \theta } \right)}} = \dfrac{{181}}{{338}}$ .

Answer 1

Hint:
Here, the value of $\tan \theta $ is negative. So, $\theta $ must lie in the second or fourth quadrant. As it is given that $\theta $ does not lie in the second quadrant, it must lie in the fourth quadrant.
Then, find the values of $\sin \theta ,\cos \theta ,\operatorname{cosec} \theta ,\sec \theta $ and $\cot \theta $ , keeping in mind the quadrant of $\theta $ .
Finally, simplify $\dfrac{{\sin \left( {360^\circ - \theta } \right) + \tan \left( {90^\circ + \theta } \right)}}{{ - \sec \left( {270 + \theta } \right) + \operatorname{cosec} \left( { - \theta } \right)}}$ and put the values of $\sin \theta ,\cos \theta ,\operatorname{cosec} \theta ,\sec \theta $ and $\cot \theta $ .

Complete step by step solution:
It is given that, $\tan \theta = - \dfrac{5}{{12}}$ and $\theta $ is not in the second quadrant.
Here, the value of $\tan \theta $ is negative. So, $\theta $ must lie in the second or fourth quadrant. As it is given that $\theta $ does not lie in the second quadrant, it must lie in the fourth quadrant.
Now, using the right-angled triangle ABC, where $\angle B = 90^\circ $ , \[AB = 5\] and \[BC = 12\] , we will find the values of $\sin \theta $ and $\cos \theta $ .

From the above figure, $AC = \sqrt {A{B^2} + B{C^2}} $
 $\therefore AC = \sqrt {{5^2} + {{12}^2}} $
             $
   = \sqrt {25 + 144} \\
   = \sqrt {169} \\
 $
             =13
Now, $\sin \theta = \dfrac{{AB}}{{AC}}$ and $\cos \theta = \dfrac{{BC}}{{AC}}$ .
 $\therefore \sin \theta = \dfrac{5}{{13}}$ and $\cos \theta = \dfrac{{12}}{{13}}$
But, $\theta $ lies in the fourth quadrant. So, $\sin \theta = - \dfrac{5}{{13}}$ and $\cos \theta = \dfrac{{12}}{{13}}$ , because in fourth quadrant, the values of sin and tan functions are negative.
Thus, $\sin \theta = - \dfrac{5}{{13}},\cos \theta = \dfrac{{12}}{{13}},\operatorname{cosec} \theta = - \dfrac{{13}}{5},\sec \theta = \dfrac{{13}}{{12}}$ and $\cot \theta = - \dfrac{{12}}{5}$ .
Now, L.H.S. $ = \dfrac{{\sin \left( {360^\circ - \theta } \right) + \tan \left( {90^\circ + \theta } \right)}}{{ - \sec \left( {270 + \theta } \right) + \operatorname{cosec} \left( { - \theta } \right)}}$
       $ = \dfrac{{ - \sin \theta - \cot \theta }}{{ - \operatorname{cosec} \theta - \operatorname{cosec} \theta }}$
Substituting the values $\sin \theta = - \dfrac{5}{{13}},\cos \theta = \dfrac{{12}}{{13}},\operatorname{cosec} \theta = - \dfrac{{13}}{5},\sec \theta = \dfrac{{13}}{{12}}$ and $\cot \theta = - \dfrac{{12}}{5}$ , we get
L.H.S. \[ = \dfrac{{ - \left( { - \dfrac{5}{{13}}} \right) - \left( { - \dfrac{{12}}{5}} \right)}}{{ - \left( { - \dfrac{{13}}{5}} \right) - \left( { - \dfrac{{13}}{5}} \right)}}\]
            \[
   = \dfrac{{\dfrac{5}{{13}} + \dfrac{{12}}{5}}}{{\dfrac{{13}}{5} + \dfrac{{13}}{5}}} \\
   = \dfrac{{\dfrac{{\left( {5 \times 5} \right) + \left( {12 \times 13} \right)}}{{13 \times 5}}}}{{2 \times \dfrac{{13}}{5}}} \\
   = \dfrac{{25 + 156}}{{26 \times 13}} \\
   = \dfrac{{181}}{{338}} \\
 \]
            = R.H.S.
Hence, proved L.H.S. = R.H.S.

Note:
Some properties of trigonometric functions used here:
 $
  \sin \theta \operatorname{cosec} \theta = 1 \\
  \cos \theta \sec \theta = 1 \\
  \tan \theta \cot \theta = 1 \\
  \sin \left( {2\pi - \theta } \right) = - \sin \theta \\
  \tan \left( {\dfrac{\pi }{2} + \theta } \right) = - \cot \theta \\
  \sec \left( {\dfrac{{3\pi }}{2} + \theta } \right) = \operatorname{cosec} \theta \\
  \operatorname{cosec} \left( { - \theta } \right) = \operatorname{cosec} \theta \\
 $