Question

Find the value of the trigonometric expression $\tan \left( {{360}^{{}^\circ \;}}-A \right)$.

Answer 1

Hint: Use the relation $\tan \left( x \right)=\dfrac{\sin \left( x \right)}{\cos \left( x \right)}$ and the expansion formula $\sin \left( s-t \right)=-\cos \left( s \right)\sin \left( t \right)+\cos \left( t \right)\sin \left( s \right)$ and$\cos \left( s-t \right)=\cos \left( s \right)\cos \left( t \right)+\sin \left( s \right)\sin \left( t \right)$, then simplify to get the required value of $\tan \left( {{360}^{{}^\circ \;}}-A \right)$.

Complete step-by-step solution -
In the question, we have to find the value of the trigonometric expression $\tan \left( {{360}^{{}^\circ \;}}-A \right)$.
Now we will use the relation $\tan \left( x \right)=\dfrac{\sin \left( x \right)}{\cos \left( x \right)}$ and we get:
$\Rightarrow \tan \left( {{360}^{{}^\circ \;}}-A \right)=\dfrac{\sin \left( -A+{{360}^{{}^\circ \;}} \right)}{\cos \left( -A+{{360}^{{}^\circ \;}} \right)}$
Next we will apply the expansion formula of $\sin \left( s-t \right)=-\cos \left( s \right)\sin \left( t \right)+\cos \left( t \right)\sin \left( s \right)$ and $\cos \left( s-t \right)=\cos \left( s \right)\cos \left( t \right)+\sin \left( s \right)\sin \left( t \right)$. Which will give us;

& \Rightarrow \dfrac{\sin \left( -A+{{360}^{{}^\circ \;}} \right)}{\cos \left( -A+{{360}^{{}^\circ \;}} \right)}=\dfrac{-\cos \left( {{360}^{{}^\circ \;}} \right)\sin \left( A \right)+\cos \left( A \right)\sin \left( {{360}^{{}^\circ \;}} \right)}{\cos \left( {{360}^{{}^\circ \;}}-A \right)} \\\ & \Rightarrow \dfrac{\sin \left( -A+{{360}^{{}^\circ \;}} \right)}{\cos \left( -A+{{360}^{{}^\circ \;}} \right)}=\dfrac{-\cos \left( {{360}^{{}^\circ \;}} \right)\sin \left( A \right)+\cos \left( A \right)\sin \left( {{360}^{{}^\circ \;}} \right)}{\cos \left( {{360}^{{}^\circ \;}} \right)\cos \left( A \right)+\sin \left( {{360}^{{}^\circ \;}} \right)\sin \left( A \right)} \\\ & \Rightarrow \dfrac{\sin \left( -A+{{360}^{{}^\circ \;}} \right)}{\cos \left( -A+{{360}^{{}^\circ \;}} \right)}=\dfrac{-\sin \left( A \right)+0}{\cos \left( A \right)+0}\,\,\,\,\,\,\,\,\,\,\,\,\because \sin \left( {{360}^{{}^\circ \;}} \right)=0,\,\,\cos \left( {{360}^{{}^\circ \;}} \right)=1 \\\ & \Rightarrow \dfrac{\sin \left( -A+{{360}^{{}^\circ \;}} \right)}{\cos \left( -A+{{360}^{{}^\circ \;}} \right)}=\dfrac{-\sin \left( A \right)}{\cos \left( A \right)} \\\ & \Rightarrow \dfrac{\sin \left( -A+{{360}^{{}^\circ \;}} \right)}{\cos \left( -A+{{360}^{{}^\circ \;}} \right)}=-\tan \left( A \right) \\\ \end{aligned}$$ So finally, we can say that $$\tan \left( {{360}^{{}^\circ \;}}-A \right)=-\tan \left( A \right)$$. So this is the required value. Note: The alternate way to solve this problem is by using the direct formula that says $$\tan \left( {{360}^{{}^\circ \;}}-\theta \right)=-\tan \left( \theta \right)$$. Here we can also use directly the formula of tangent of difference of two angles like $\tan(A-B) = \dfrac{\tan A - \tan B}{1+ \tan A \tan B}$ where A = ${360}^\circ$ and B = A. For Such type of trigonometric problem we have many approaches and we get the same answer if we apply correctly all the trigonometric formulas and identities.