Question

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

Answer 1

Hint:In such questions, we prove them by either making the left hand side that is L.H.S. or by making the right hand side that is R.H.S. equal to the other in order to prove the proof that has been asked.We make use of trigonometric relations of sum and difference of angles to obtain the results.

Complete step-by-step answer:
Now, in such questions, if one tries to simplify the right hand side that is R.H.S., then first thing that is to be done is to convert the tan function in terms of sin and cos functions and that is done by using the following relations
$\tan x=\dfrac{\sin x}{\cos x}$
Now, this is the result that would be used to prove the proof mentioned in this question as using this identity, we would convert the left hand side that is L.H.S. or the right hand side that is R.H.S. to make either of them equal to the other.

As mentioned in the question, we have to prove the given expression that is
$\dfrac{\cos {{8}^{\circ }}-\sin {{8}^{\circ }}}{\cos {{8}^{\circ }}+\sin {{8}^{\circ }}}=\tan {{37}^{\circ }}$ .
Now, we will start with the left hand side that is L.H.S. as follows
$=\dfrac{\cos {{8}^{\circ }}-\sin {{8}^{\circ }}}{\cos {{8}^{\circ }}+\sin {{8}^{\circ }}}$
Now, we will divide every term in the numerator as well as the denominator with $\cos {{9}^{\circ }}$ to get to the solution as follows

& =\dfrac{1-\dfrac{\sin {{8}^{\circ }}}{\cos {{8}^{\circ }}}}{1+\dfrac{\sin {{8}^{\circ }}}{\cos {{8}^{\circ }}}} \\\ & =\dfrac{1-\tan {{8}^{\circ }}}{1+\tan {{8}^{\circ }}} \\\ \end{aligned}$$ (Using the identities that are mentioned in the hint) Now, we know that the value of $$\tan {{45}^{\circ }}=1$$ Now, using the relation that has been mentioned above, we can write the above expression as follows $$\begin{aligned} & =\dfrac{\tan {{45}^{\circ }}-\tan {{8}^{\circ }}}{\tan {{45}^{\circ }}+\tan {{8}^{\circ }}} \\\ & =\dfrac{\dfrac{\sin {{45}^{\circ }}}{\cos {{45}^{\circ }}}-\dfrac{\sin {{8}^{\circ }}}{\cos {{8}^{\circ }}}}{\dfrac{\sin {{45}^{\circ }}}{\cos {{45}^{\circ }}}+\dfrac{\sin {{8}^{\circ }}}{\cos {{8}^{\circ }}}} \\\ & =\dfrac{\sin {{45}^{\circ }}\cdot \cos {{8}^{\circ }}-\sin {{8}^{\circ }}\cdot \cos {{45}^{\circ }}}{\sin {{45}^{\circ }}\cdot \cos {{8}^{\circ }}+\sin {{8}^{\circ }}\cdot \cos {{45}^{\circ }}} \\\ & =\dfrac{\sin ({{45}^{\circ }}-{{8}^{\circ }})}{\sin ({{45}^{\circ }}+{{8}^{\circ }})} \\\ & =\dfrac{\sin ({{37}^{\circ }})}{\sin ({{53}^{\circ }})} \\\ \end{aligned}$$ Now, we also know that the value of $$\sin \left( {{90}^{\circ }}-x \right)=\cos x$$ So, we can write as follows $$\begin{aligned} & =\dfrac{\sin ({{37}^{\circ }})}{\sin ({{53}^{\circ }})} \\\ & =\dfrac{\sin ({{37}^{\circ }})}{\sin ({{90}^{\circ }}-{{37}^{\circ }})} \\\ & =\dfrac{\sin ({{37}^{\circ }})}{\cos ({{37}^{\circ }})} \\\ & =\tan {{37}^{\circ }} \\\ \end{aligned}$$ Now, as the right hand side that is R.H.S. is equal to the left hand side that is L.H.S., hence, the expression has been proved. Note:Another method of attempting this question is by converting the right hand side that is R.H.S. to the left hand side that is L.H.S. by using the relations that are given in the hint. Through this method also, we could get to the correct answer and hence, we would be able to prove the required proof.