Question

Using the trigonometric identity for the tangent of sum of two angles, prove that
$\dfrac{\cos \left( {{8}^{\circ }} \right)-\sin \left( {{8}^{\circ }} \right)}{\cos \left( {{8}^{\circ }} \right)+\sin \left( {{8}^{\circ }} \right)}=\tan ({{37}^{\circ }})$

Answer 1

Hint: In this question, the LHS has the terms consisting of cosine and sine terms and the RHS consists of a tangent term. Therefore, we should try to convert both sides in terms of tan and then use the formula for tangent of difference of two angles to obtain the required answer.

Complete step-by-step answer:
In this equation, we should try to convert both sides in terms of the same trigonometric ratio i.e. tan. Therefore, we should use the definition of tan in terms of sin and cos which is
$\tan \left( \theta \right)=\dfrac{\sin \left( \theta \right)}{\cos \left( \theta \right)}..................(1.1)$
Therefore as $\cos ({{8}^{\circ }})\ne 0$ , we can divide both numerator and denominator of LHS by $\cos ({{8}^{\circ }})$ and use equation (1.1) to obtain
$\dfrac{\cos \left( {{8}^{\circ }} \right)-\sin \left( {{8}^{\circ }} \right)}{\cos \left( {{8}^{\circ }} \right)+\sin \left( {{8}^{\circ }} \right)}=\dfrac{\dfrac{\cos \left( {{8}^{\circ }} \right)}{\cos \left( {{8}^{\circ }} \right)}-\dfrac{\sin \left( {{8}^{\circ }} \right)}{\cos \left( {{8}^{\circ }} \right)}}{\dfrac{\cos \left( {{8}^{\circ }} \right)}{\cos \left( {{8}^{\circ }} \right)}+\dfrac{\sin \left( {{8}^{\circ }} \right)}{\cos \left( {{8}^{\circ }} \right)}}=\dfrac{1-\tan \left( {{8}^{\circ }} \right)}{1+\tan \left( {{8}^{\circ }} \right)}..........(1.2)$
As the tangent of ${{45}^{\circ }}$ is given by $\tan ({{45}^{\circ }})=1$. Therefore, replacing 1 in equation (1.2) by $\tan ({{45}^{\circ }})$ to obtain
$\dfrac{\cos \left( {{8}^{\circ }} \right)-\sin \left( {{8}^{\circ }} \right)}{\cos \left( {{8}^{\circ }} \right)+\sin \left( {{8}^{\circ }} \right)}=\dfrac{1-\tan \left( {{8}^{\circ }} \right)}{1+\tan \left( {{8}^{\circ }} \right)}=\dfrac{\tan \left( {{45}^{\circ }} \right)-\tan \left( {{8}^{\circ }} \right)}{1+\tan \left( {{45}^{\circ }} \right)\tan \left( {{8}^{\circ }} \right)}..........(1.3)$
Now, the tangent of a difference of two angles is given by
$\tan (a-b)=\dfrac{\tan (a)-\tan (b)}{1+\tan (a)\tan (b)}...........(1.4)$
Therefore, taking $a={{45}^{\circ }}$ and $b={{8}^{\circ }}$ in equation (1.4), we obtain
$\tan ({{45}^{\circ }}-{{8}^{\circ }})=\dfrac{\tan ({{45}^{\circ }})-\tan ({{8}^{\circ }})}{1+\tan ({{45}^{\circ }})\tan ({{8}^{\circ }})}...........(1.5)$
Thus, by using equation (1.3) and (1.5), we can write the equation as
$\dfrac{\cos \left( {{8}^{\circ }} \right)-\sin \left( {{8}^{\circ }} \right)}{\cos \left( {{8}^{\circ }} \right)+\sin \left( {{8}^{\circ }} \right)}=\dfrac{\tan \left( {{45}^{\circ }} \right)-\tan \left( {{8}^{\circ }} \right)}{1+\tan \left( {{45}^{\circ }} \right)\tan \left( {{8}^{\circ }} \right)}=\tan \left( {{45}^{\circ }}-{{8}^{\circ }} \right)=\tan ({{37}^{\circ }})........(1.6)$

Thus, from equation (1.6), we get
$\dfrac{\cos \left( {{8}^{\circ }} \right)-\sin \left( {{8}^{\circ }} \right)}{\cos \left( {{8}^{\circ }} \right)+\sin \left( {{8}^{\circ }} \right)}=\tan ({{37}^{\circ }})$
Which is exactly what we wanted to prove.

Note: We should note that when we divide $\cos ({{8}^{\circ }})$ it is necessary to state that it is not equal to zero as otherwise the resulting equation will not be valid as division by zero is undefined.