Question

If $ \sec \theta + \tan \theta = \dfrac{1}{5} $ , then the value of $ \sin \theta $ is:
 $
  (1)\,\,\,\dfrac{{12}}{{13}} \\
  (2)\,\, - \dfrac{{12}}{{13}} \\
  (3)\,\, \pm \dfrac{{12}}{{13}} \\
  (4)\,\,\dfrac{5}{{12}} \\
  $

Answer 1

Hint: To find required value we use trigonometry identity and then expanding it using algebraic identity and then simplifying it by substituting given value to form another equation which on solving with given equation gives out value of $ \tan \theta \,\,and\,\,\sec \theta $ and then using these we can find value of required $ \sin \theta $ .
Formulas used: $ {\sec ^2}\theta - {\tan ^2}\theta = 1 $

Complete step-by-step answer:
To find the required value of $ \sin \theta $ . We consider trigonometric identity:
 $ {\sec ^2}\theta - {\tan ^2}\theta = 1 $
Simplifying left hand side using algebraic identity. We have,
 $ \left( {\sec \theta + \tan \theta } \right)\left( {\sec \theta - \tan \theta } \right) = 1 $
But it is given $ \sec \theta + \tan \theta = \dfrac{1}{5} $ ……….(i)
Substituting value in above equation. We have
 $
  \dfrac{1}{5}\left( {\sec \theta - \tan \theta } \right) = 1 \\
   \Rightarrow \sec \theta - \tan \theta = 5.....................(ii) \;
  $
Adding equation (i) and (ii) formed above
 $
  2\sec \theta = \dfrac{1}{5} + 5 \\
   \Rightarrow 2\sec \theta = \dfrac{{1 + 25}}{5} \\
   \Rightarrow 2\sec \theta = \dfrac{{26}}{5} \\
   \Rightarrow \sec \theta = \dfrac{{26}}{5} \times \dfrac{1}{2} \\
   \Rightarrow \sec \theta = \dfrac{{13}}{5} \;
  $
Substituting $ \sec \theta = \dfrac{{13}}{5} $ in equation (i). We have
 $
  \dfrac{{13}}{5} + \tan \theta = \dfrac{1}{5} \\
   \Rightarrow \tan \theta = \dfrac{1}{5} - \dfrac{{13}}{5} \\
   \Rightarrow \tan \theta = \dfrac{{1 - 13}}{5} \\
   \Rightarrow \tan \theta = - \dfrac{{12}}{5} \;
  $
Now, dividing value $ \tan \theta \,\,with\,\,value\,\,of\,\,\sec \theta \,\,calculated\,\,in\,\,above. $
 $
 \Rightarrow \dfrac{{\tan \theta }}{{\sec \theta }} = \dfrac{{ - \dfrac{{12}}{5}}}{{\dfrac{{13}}{5}}} \\
   \Rightarrow \dfrac{{\dfrac{{\sin \theta }}{{\cos \theta }}}}{{\dfrac{1}{{\cos \theta }}}} = - \dfrac{{12}}{5} \times \dfrac{5}{{13}} \\
   \Rightarrow \sin \theta = - \dfrac{{12}}{{13}} \;
  $
Therefore, from above we see that the value of $ \sin \theta \,\,is\,\, - \dfrac{{12}}{{13}} $ .
So, the correct answer is “ $ \dfrac{{-12}}{{13}} $ ”.

Note: We can also find the required value from a given function in another way. In this we first convert given trigonometric function in term of $ \sin \theta \,\,and\,\cos \theta $ and then squaring both side and converting in term of $ \sin \theta $ to form quadratic equation and then solving quadratic so formed to get value of $ \sin \theta $ and hence required solution of given problem.