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Question: If \( \sec \theta + \tan \theta = \dfrac{1}{5} \) , then the value of \( \sin \theta \) is: \( ...

If sec⁡θ+tan⁡θ=15\sec \theta + \tan \theta = \dfrac{1}{5} , then the value of sin⁡θ\sin \theta is:
(1)   1213 (2)  −1213 (3)  ±1213 (4)  512  (1)\,\,\,\dfrac{{12}}{{13}} \\\ (2)\,\, - \dfrac{{12}}{{13}} \\\ (3)\,\, \pm \dfrac{{12}}{{13}} \\\ (4)\,\,\dfrac{5}{{12}} \\\

Explanation

Solution

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⁡θ  and  sec⁡θ\tan \theta \,\,and\,\,\sec \theta and then using these we can find value of required sin⁡θ\sin \theta .
Formulas used: sec⁡2θ−tan⁡2θ=1{\sec ^2}\theta - {\tan ^2}\theta = 1

Complete step-by-step answer :
To find the required value of sin⁡θ\sin \theta . We consider trigonometric identity:
sec⁡2θ−tan⁡2θ=1{\sec ^2}\theta - {\tan ^2}\theta = 1
Simplifying left hand side using algebraic identity. We have,
(sec⁡θ+tan⁡θ)(sec⁡θ−tan⁡θ)=1\left( {\sec \theta + \tan \theta } \right)\left( {\sec \theta - \tan \theta } \right) = 1
But it is given sec⁡θ+tan⁡θ=15\sec \theta + \tan \theta = \dfrac{1}{5} ……….(i)
Substituting value in above equation. We have
15(sec⁡θ−tan⁡θ)=1 ⇒sec⁡θ−tan⁡θ=5.....................(ii)   \dfrac{1}{5}\left( {\sec \theta - \tan \theta } \right) = 1 \\\ \Rightarrow \sec \theta - \tan \theta = 5.....................(ii) \;
Adding equation (i) and (ii) formed above
2sec⁡θ=15+5 ⇒2sec⁡θ=1+255 ⇒2sec⁡θ=265 ⇒sec⁡θ=265×12 ⇒sec⁡θ=135   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⁡θ=135\sec \theta = \dfrac{{13}}{5} in equation (i). We have
135+tan⁡θ=15 ⇒tan⁡θ=15−135 ⇒tan⁡θ=1−135 ⇒tan⁡θ=−125   \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⁡θ  with  value  of  sec⁡θ  calculated  in  above.\tan \theta \,\,with\,\,value\,\,of\,\,\sec \theta \,\,calculated\,\,in\,\,above.
⇒tan⁡θsec⁡θ=−125135 ⇒sin⁡θcos⁡θ1cos⁡θ=−125×513 ⇒sin⁡θ=−1213   \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⁡θ  is  −1213\sin \theta \,\,is\,\, - \dfrac{{12}}{{13}} .
So, the correct answer is “ −1213\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⁡θ  and cos⁡θ\sin \theta \,\,and\,\cos \theta and then squaring both side and converting in term of sin⁡θ\sin \theta to form quadratic equation and then solving quadratic so formed to get value of sin⁡θ\sin \theta and hence required solution of given problem.