Question

What is the equivalent value of
$\dfrac{3+\cot {{76}^{\circ }}\cot {{16}^{\circ }}}{\cot {{76}^{\circ }}+\cot {{16}^{\circ }}}$
(a) $\tan {{44}^{\circ }}$
(b) $\cot {{46}^{\circ }}$
(c) $\tan {{2}^{\circ }}$
(d) $\tan {{46}^{\circ }}$

Answer 1

Hint: In this question, we are given the expression in terms of cot of ${{76}^{\circ }}$ and ${{16}^{\circ }}$. We should first try to convert all the trigonometric ratios in terms of sine and cosine and then use the trigonometric identities to simplify the given expression.

Complete step by step solution:
The expression given in the question is
$\dfrac{3+\cot {{76}^{\circ }}\cot {{16}^{\circ }}}{\cot {{76}^{\circ }}+\cot {{16}^{\circ }}}...............(1.1)$
As we know that cot is given by
$\cot \theta =\dfrac{\cos \theta }{\sin \theta }$
for any angle $\theta$, we can use it to rewrite equation (1.1) as
$\dfrac{3+\cot {{76}^{\circ }}\cot {{16}^{\circ }}}{\cot {{76}^{\circ }}+\cot {{16}^{\circ }}}=\dfrac{2+1+\dfrac{\cos {{76}^{\circ }}}{\sin {{76}^{\circ }}}\times \dfrac{\cos {{16}^{\circ }}}{\sin {{16}^{\circ }}}}{\dfrac{\cos {{76}^{\circ }}}{\sin {{76}^{\circ }}}+\dfrac{\cos {{16}^{\circ }}}{\sin {{16}^{\circ }}}}=\dfrac{2+\dfrac{\sin {{76}^{\circ }}\sin {{16}^{\circ }}+\cos {{76}^{\circ }}\cos {{16}^{\circ }}}{\sin {{76}^{\circ }}\sin {{16}^{\circ }}}}{\dfrac{\cos {{76}^{\circ }}\sin {{16}^{\circ }}+\cos {{16}^{\circ }}\sin {{76}^{\circ }}}{\sin {{76}^{\circ }}\sin {{16}^{\circ }}}}................(1.2)$
Now, from the trigonometric identities of cosine and sine of sum of angles, we have
$\begin{aligned} & \cos (a-b)=\cos a\cos b+\sin a\sin b \\\ & \sin (a+b)=\sin a\cos b+\cos a\sin b.................(1.3) \\\ \end{aligned}$
Now, taking $a={{76}^{\circ }}$ and $b={{16}^{\circ }}$ in equation (1.3) and using it in equation (1.2), we get

& \dfrac{3+\cot {{76}^{\circ }}\cot {{16}^{\circ }}}{\cot {{76}^{\circ }}+\cot {{16}^{\circ }}}=\dfrac{2+\dfrac{\sin {{76}^{\circ }}\sin {{16}^{\circ }}+\cos {{76}^{\circ }}\cos {{16}^{\circ }}}{\sin {{76}^{\circ }}\sin {{16}^{\circ }}}}{\dfrac{\cos {{76}^{\circ }}\sin {{16}^{\circ }}+\cos {{16}^{\circ }}\sin {{76}^{\circ }}}{\sin {{76}^{\circ }}\sin {{16}^{\circ }}}}=\dfrac{2+\dfrac{\cos \left( {{76}^{\circ }}-{{16}^{\circ }} \right)}{\sin {{76}^{\circ }}\sin {{16}^{\circ }}}}{\dfrac{\sin \left( {{76}^{\circ }}+{{16}^{\circ }} \right)}{\sin {{76}^{\circ }}\sin {{16}^{\circ }}}} \\\ & =\dfrac{\dfrac{2\sin {{76}^{\circ }}\sin {{16}^{\circ }}+\cos \left( {{60}^{\circ }} \right)}{\sin {{76}^{\circ }}\sin {{16}^{\circ }}}}{\dfrac{\sin \left( {{92}^{\circ }} \right)}{\sin {{76}^{\circ }}\sin {{16}^{\circ }}}}................(1.3) \\\ \end{aligned}$$ Now, we can cancel out $$\sin {{76}^{\circ }}\sin {{16}^{\circ }}$$ from numerator and denominator to get $$\begin{aligned} & \dfrac{3+\cot {{76}^{\circ }}\cot {{16}^{\circ }}}{\cot {{76}^{\circ }}+\cot {{16}^{\circ }}} \\\ & =\dfrac{2\sin {{76}^{\circ }}\sin {{16}^{\circ }}+\cos \left( {{60}^{\circ }} \right)}{\sin \left( {{92}^{\circ }} \right)}................(1.4) \\\ \end{aligned}$$ Now, we can use the trigonometric identity $2\sin \left( a \right)\sin \left( b \right)=-\cos \left( b+a \right)+\cos \left( b-a \right)$ If we take$b={{76}^{\circ }}$ and $a={{16}^{\circ }}$ to obtain $\begin{aligned} & 2\sin \left( {{76}^{\circ }} \right)\sin \left( {{16}^{\circ }} \right)=-\cos \left( {{76}^{\circ }}+{{16}^{\circ }} \right)+\cos \left( {{76}^{\circ }}-{{16}^{\circ }} \right) \\\ & =-\cos \left( {{92}^{\circ }} \right)+\cos \left( {{60}^{\circ }} \right)...............(1.5) \\\ \end{aligned}$ We can now use equation (1.5) in equation (1.4) to obtain $$\begin{aligned} & \dfrac{3+\cot {{76}^{\circ }}\cot {{16}^{\circ }}}{\cot {{76}^{\circ }}+\cot {{16}^{\circ }}} \\\ & =\dfrac{2\sin {{76}^{\circ }}\sin {{16}^{\circ }}+\cos \left( {{60}^{\circ }} \right)}{\sin \left( {{92}^{\circ }} \right)}=\dfrac{-\cos \left( {{92}^{\circ }} \right)+\cos \left( {{60}^{\circ }} \right)+\cos \left( {{60}^{\circ }} \right)}{\sin \left( {{92}^{\circ }} \right)} \\\ & =\dfrac{-\cos \left( {{92}^{\circ }} \right)+\dfrac{1}{2}+\dfrac{1}{2}}{\sin \left( {{92}^{\circ }} \right)}=\dfrac{1-\cos \left( {{92}^{\circ }} \right)}{\sin \left( {{92}^{\circ }} \right)}................(1.6) \\\ \end{aligned}$$ Again, we know the formula $2{{\sin }^{2}}\theta =1-\cos \left( 2\theta \right)................(1.7)$ And $\sin \left( 2\theta \right)=2\sin \theta \cos \theta ...............(1.8)$ Thus, taking $\theta ={{46}^{\circ }}\Rightarrow 2\theta ={{92}^{\circ }}$ in equation (1.7) and (1.8) and using it in (1.6), we obtain $$\dfrac{3+\cot {{76}^{\circ }}\cot {{16}^{\circ }}}{\cot {{76}^{\circ }}+\cot {{16}^{\circ }}}=\dfrac{1-\cos \left( {{92}^{\circ }} \right)}{\sin \left( {{92}^{\circ }} \right)}=\dfrac{2{{\sin }^{2}}\left( {{46}^{\circ }} \right)}{2\sin \left( {{46}^{\circ }} \right)\cos \left( {{46}^{\circ }} \right)}................(1.9)$$ Cancelling $\sin \left( {{46}^{\circ }} \right)$ from numerator and denominator, we obtain $$\dfrac{3+\cot {{76}^{\circ }}\cot {{16}^{\circ }}}{\cot {{76}^{\circ }}+\cot {{16}^{\circ }}}=\dfrac{1-\cos \left( {{92}^{\circ }} \right)}{\sin \left( {{92}^{\circ }} \right)}=\dfrac{\sin \left( {{46}^{\circ }} \right)}{\cos \left( {{46}^{\circ }} \right)}=\tan \left( {{46}^{\circ }} \right)$$ Which matches option (d) of the given question. Hence (d) is the correct answer to this question. Note: We should note that we have converted everything in terms of sin and cos to solve this question. However, one can also convert the terms into other trigonometric ratios such as tan and cot and use their relations to solve this question. However, the final answer will remain the same in all the methods.