Question

How do you write the expression \[~\dfrac{\tan {{325}^{\circ }}-\tan {{86}^{\circ }}}{1+\tan {{325}^{\circ }}\tan {{86}^{\circ }}}\] as the sine, cosine, or tangent of an angle?

Answer 1

Hint: Take the angle A as \[{{325}^{\circ }}\] and the angle B as \[{{86}^{\circ }}\]. Then use the formula \[\tan (A-B)=~\dfrac{\tan A-\tan B}{1+\tan A\tan B}\], to get the value of \[\tan (A-B)\]. Convert tanget to sine and cosine using the formula $\tan \theta =\dfrac{\sin \theta }{\cos \theta }$.

Complete step by step solution:
Considering the expression \[~\dfrac{\tan {{325}^{\circ }}-\tan {{86}^{\circ }}}{1+\tan {{325}^{\circ }}\tan {{86}^{\circ }}}\]
Clearly, the given expression is in the form of \[\tan (A-B)\], as \[\tan (A-B)=~\dfrac{\tan A-\tan B}{1+\tan A\tan B}\]
Here, A =\[{{325}^{\circ }}\] and B =\[{{86}^{\circ }}\]
Therefore, \[~~\dfrac{\tan {{325}^{\circ }}-\tan {{86}^{\circ }}}{1+\tan {{325}^{\circ }}\tan {{86}^{\circ }}}=\tan ({{325}^{\circ }}-{{86}^{\circ }})=\tan {{239}^{\circ }}\]
To express the above expression in sine and cosine of an angle; we know that$\tan \theta =\dfrac{\sin \theta }{\cos \theta }$
So, $\tan {{239}^{\circ }}$ can be written as $\tan {{239}^{\circ }}=\dfrac{\sin {{239}^{\circ }}}{\cos {{239}^{\circ }}}$.
This is the required solution of the given question.

Note: Recognizing the expression \[~\dfrac{\tan {{325}^{\circ }}-\tan {{86}^{\circ }}}{1+\tan {{325}^{\circ }}\tan {{86}^{\circ }}}\] as in the from of \[\tan (A-B)=~\dfrac{\tan A-\tan B}{1+\tan A\tan B}\] should be the first approach for solving this question. Tangent of an angle can be written in terms of sine and cosine using the formula $\tan \theta =\dfrac{\sin \theta }{\cos \theta }$.