Question

Prove that: $\dfrac{\cos 8{}^\circ -\sin 8{}^\circ }{\cos 8{}^\circ +\sin 8{}^\circ }=\tan 37{}^\circ$

Answer 1

Hint: We will first start by dividing the numerator and denominator by $\cos 8{}^\circ $. Then, we will use the trigonometric identity that $\tan \left( A-B \right)=\dfrac{\tan A-\tan B}{1+\tan A\tan B}$ to prove the given result.

Complete step-by-step answer:
Now, we have to prove that $\dfrac{\cos 8{}^\circ -\sin 8{}^\circ }{\cos 8{}^\circ +\sin 8{}^\circ }=\tan 37{}^\circ $.
Now, we will take LHS given to us as,
$\dfrac{\cos 8{}^\circ -\sin 8{}^\circ }{\cos 8{}^\circ +\sin 8{}^\circ }$
Now, we divide the both numerator and denominator with $\cos 8{}^\circ $. So, we have,
$\Rightarrow \dfrac{\dfrac{\cos 8{}^\circ -\sin 8{}^\circ }{\cos 8{}^\circ }}{\dfrac{\cos 8{}^\circ +\sin 8{}^\circ }{\cos 8{}^\circ }}$
Now, we know that $\dfrac{\sin A}{\cos A}=\tan A$
$\Rightarrow \dfrac{1-\tan 8{}^\circ }{1+\tan 8{}^\circ }$
Now, we know that $1=\tan 45{}^\circ $
$\Rightarrow \dfrac{\tan 45{}^\circ -\tan 8{}^\circ }{1+\tan 45{}^\circ \times \tan 8{}^\circ }$
Now, we know that $\dfrac{\tan A-\tan B}{1+\tan A\tan B}=\tan \left( A-B \right)$
$\begin{align}
  & \Rightarrow \tan \left( 45{}^\circ -8{}^\circ \right) \\
 & =\tan 37{}^\circ \\
\end{align}$
Since, we have LHS = RHS.
Hence Proved.

Note: It is important to note that how we have used the fact that $\tan \left( A-B \right)=\dfrac{\tan A-\tan B}{1+\tan A\tan B}$ to convert the expression $\dfrac{1-\tan 8{}^\circ }{1+\tan 8{}^\circ }$ to $\dfrac{\tan 45{}^\circ -\tan 8{}^\circ }{1+\tan 45{}^\circ \times \tan 8{}^\circ }$. Also, in this conversion we have deliberately left the 1 in denominator unchanged. So, that the formula can be applied successfully.