Question

Find the general value of the trigonometric term: $3({\sec ^2}\theta + {\tan ^2}\theta ) = 5$ respectively.
A. $\pi$
B. $n\pi \pm \dfrac{\pi }{6}$
C. $n\pi \mp \dfrac{\pi }{6}$
D. $n\pi \pm \dfrac{6}{\pi }$

Answer 1

The given problem revolves around the concepts of trigonometric equations. So, we will use the definition of trigonometric equations and its identities. Here, we have converted the given terms in ‘sine’ and ‘cosine’ forms and by using the certain identities and values at respective angles (both, in degree and radian) say, ${\sin ^2}\theta + {\cos ^2}\theta = 1$, the desired solution can be obtained.

Complete step by step answer:
Since, we have given the equation
$\Rightarrow 3({\sec ^2}\theta + {\tan ^2}\theta ) = 5$
As a result, rearranging the terms, we get
$\Rightarrow {\sec ^2}\theta + {\tan ^2}\theta = \dfrac{5}{3}$
Converting the certain trigonometric terms ‘sec’ and ‘tan’ into ‘sin’, ‘cos’ terms, we get

\Rightarrow \dfrac{{1 + {{\sin }^2}\theta }}{{{{\cos }^2}\theta }} = \dfrac{5}{3} \\\ $$ Multiplying $3{\text{ and }}{\cos ^2}\theta $ to both sides, we get $$ \Rightarrow 3(1 + {\sin ^2}\theta ) = 5{\cos ^2}\theta $$ … (i) Now, since we know that the trigonometric identity for ‘sine’ and ‘cosine’ terms is ${\sin ^2}\theta + {\cos ^2}\theta = 1$, As a result, substituting the value of ‘cosine’ term from an identity, in equation (i), we get $$\Rightarrow 3(2 - {\cos ^2}\theta ) = 5{\cos ^2}\theta \\\ \Rightarrow 6 - 3{\cos ^2}\theta = 5{\cos ^2}\theta \\\ $$ Taking constant on one side and remaining terms on other side, we get $$ \Rightarrow 8{\cos ^2}\theta = 6$$ Dividing $8$ to both sides, we get, $$ \Rightarrow {\cos ^2}\theta = \dfrac{6}{8} = \dfrac{3}{4}$$ Taking square roots on both sides, we get $$ \Rightarrow \cos \theta = \pm \dfrac{{\sqrt 3 }}{2}$$ Where, square roots can be both positive as well as negative one, Therefore, ‘$ \pm $’ sign exists! As a result, we also know that, the value $$\dfrac{{\sqrt 3 }}{2}$$ of cosine in trigonometric condition exists only when the angle theta i.e. $\theta = {30^ \circ } = {\dfrac{\pi }{6}\,{rad}}$ The equation becomes, $$\cos \theta = \pm \cos \dfrac{\pi }{6}$$ Hence, in general we can write the value of $$\theta $$in terms of following, $$ \therefore \theta = n\pi \pm \cos \dfrac{\pi }{6}$$ **Hence, option B is correct.** **Note:** One must know how to convert the ‘tan’, ‘cot’, ‘sec’ and ‘cosec’ terms in trigonometric identities respectively considering their formulae such as, ${\sin ^2}\theta + {\cos ^2}\theta = 1{\text{, }}1 + {\tan ^2}\theta = {\sec ^2}\theta {\text{ and }}1 + {\cot ^2}\theta = \cos e{c^2}\theta $ so as to distinguish the solution accurately. Also, we should know all the required values of standard angles say, $${0^o},{30^o},{45^o},{60^o},{90^o},{180^o},{270^o},{360^o}$$ respectively for each trigonometric term such as$\sin ,\cos ,\tan ,\cot ,\sec ,\cos ec$, etc. We should take care of the calculations to convert the angles from ‘degrees’ to ‘radian’ form say, ${30^ \circ } = {30^ \circ } \times \dfrac{\pi }{{180}} = {\dfrac{\pi }{6}^{radian}}$ so as to be sure of our final answer.