Question

Given the value of \[sec\;\theta = \dfrac{5}{4}\], then find the values of \[tan\left( {\dfrac{\theta }{2}} \right)\] and \[tan\;\theta \].

Answer 1

Hint: Look for a pattern that suggests an algebraic property, such as the difference of squares or a factoring opportunity. Substitute the trigonometric expression with a single variable if required. Solve the equation the same way an algebraic equation would be solved. Substitute the trigonometric expression back in for the variable in the resulting expressions.

Complete step-by-step answer:
Use the definition of secant to find the known sides of the unit circle right triangle. The quadrant determines the sign on each of the values.
\[sec\left( \theta \right) = \dfrac{{hypotenuse}}{{adjacent}}\]
Find the opposite side of the unit circle triangle. Since the adjacent side and hypotenuse are known, use the Pythagorean theorem to find the remaining side.
\[Opposite = \sqrt {hypotenus{e^2} - adjacen{t^2}} \]
Replace the known values in the equation.
\[ \Rightarrow Opposite = \sqrt {{{\left( 5 \right)}^2} - {{\left( 4 \right)}^2}} \]
Simplify by squaring both 5 and 4 and then subtracting them.
\[ \Rightarrow Opposite = \sqrt 9 = 3\]
\[\tan \left( \theta \right) = \dfrac{{opposite}}{{adjacent}}\]
By substituting the value of opposite and adjacent
\[\tan \left( \theta \right) = \dfrac{3}{4}\]
Now we will use an identity:-
\[\tan \theta = \dfrac{{2\tan \dfrac{\theta }{2}}}{{1 - {{\tan }^2}\dfrac{\theta }{2}}} = \dfrac{3}{4}\]
Now we will convert the equation into quadratic equation and solve the equation for the value of $\tan \dfrac{\theta }{2}$
\[ \Rightarrow 2\tan \dfrac{\theta }{2} = \dfrac{3}{4}\left( {1 - {{\tan }^2}\dfrac{\theta }{2}} \right)\]
Taking all the expressions on one side.
\[ \Rightarrow 3{\tan ^2}\dfrac{\theta }{2} + 8\tan \dfrac{\theta }{2} - 3 = 0\]
Factorize the expression:
\[ \Rightarrow 3{\tan ^2}\dfrac{\theta }{2} + 9\tan \dfrac{\theta }{2} - \tan \dfrac{\theta }{2} - 3 = 0\]
\[ \Rightarrow 3\tan \dfrac{\theta }{2}\left( {\tan \dfrac{\theta }{2} + 3} \right) - \left( {\tan \dfrac{\theta }{2} + 3} \right) = 0\]
\[ \Rightarrow \left( {3\tan \dfrac{\theta }{2} - 1} \right)\left( {\tan \dfrac{\theta }{2} + 3} \right) = 0\]
\[ \Rightarrow \tan \dfrac{\theta }{2} = - 3,\dfrac{1}{3}\]

Note: When solving linear trigonometric equations, we can use algebraic techniques just as we do solving algebraic equations. Look for patterns, like the difference of squares, quadratic form, or an expression that lends itself well to substitution.
Equations involving a single trigonometric function can be solved or verified using the unit circle.
Many equations appear quadratic in form. We can use substitution to make the equation appear simpler, and then use the same techniques we use solving an algebraic quadratic: factoring, the quadratic formula, etc.
We can also use the identities to solve trigonometric equations.