Question

Prove the following trigonometric function:
$\cot \left( {{105}^{\circ }} \right)-\tan \left( {{105}^{\circ }} \right)=2\sqrt{3}$

Answer 1

Hint:In this question, we find that the left hand side consists of terms involving tan and cot. We can use the formulas for finding the tan and cot of sum of an angle and ${{90}^{\circ }}$. Thereafter, we can use the formula for finding the tangent of twice an angle to obtain the required value given in the right hand side.

Complete step-by-step answer:
In this question, in the LHS we are given the values in tan and cot. Now, we note that
${{105}^{\circ }}={{90}^{\circ }}+{{15}^{\circ }}$
Therefore, we should use the following trigonometric identities
$\begin{aligned} & \sin \left( {{90}^{\circ }}+\theta \right)=\cos \left( \theta \right) \\\ & \cos \left( {{90}^{\circ }}+\theta \right)=-\sin \left( \theta \right) \\\ \end{aligned}$
And therefore
$\begin{aligned} & \cot \left( {{105}^{\circ }} \right)=\dfrac{\cos \left( {{105}^{\circ }} \right)}{\sin \left( {{105}^{\circ }} \right)}=\dfrac{\cos \left( {{90}^{\circ }}+{{15}^{\circ }} \right)}{\sin \left( {{90}^{\circ }}+{{15}^{\circ }} \right)} \\\ & =\dfrac{-\sin \left( {{15}^{\circ }} \right)}{\cos \left( {{15}^{\circ }} \right)}=-\tan \left( {{15}^{\circ }} \right)............(1.1) \\\ \end{aligned}$
And for tan
$\begin{aligned} & \tan \left( {{105}^{\circ }} \right)=\dfrac{\sin \left( {{105}^{\circ }} \right)}{\cos \left( {{105}^{\circ }} \right)}=\dfrac{\sin \left( {{90}^{\circ }}+{{15}^{\circ }} \right)}{\cos \left( {{90}^{\circ }}+{{15}^{\circ }} \right)} \\\ & =\dfrac{\cos \left( {{15}^{\circ }} \right)}{-\sin \left( {{15}^{\circ }} \right)}=\dfrac{-1}{\tan \left( {{15}^{\circ }} \right)}................(1.2) \\\ \end{aligned}$

Therefore, using (1.1) and (1.2) in the expression given in the question, we obtain
$\begin{aligned}
& \cot \left( {{105}^{\circ }} \right)-\tan \left( {{105}^{\circ }} \right)=-\tan \left( {{15}^{\circ }} \right)-\left( \dfrac{-1}{\tan \left( {{15}^{\circ }} \right)} \right) \\

& =\dfrac{1-{{\tan }^{2}}\left( {{15}^{\circ }} \right)}{\tan ({{15}^{\circ }})}
\end{aligned}$Multiplying and dividing the expression by 2 and taking the numerator to the denominator part we get,$=\dfrac{2}{\left( \dfrac{2\tan ({{15}^{\circ }})}{1-{{\tan }^{2}}({{15}^{\circ }})} \right)}..................(1.3)$Now, the tangent for twice an angle is given by the formula$\tan (2\theta )=\dfrac{2\tan (\theta )}{1-{{\tan }^{2}}(\theta )}...............(1.4) Therefore, using equation (1.4) in the denominator of equation (1.3), we obtain $$\cot \left( {{105}^{\circ }} \right)-\tan \left( {{105}^{\circ }} \right)=\dfrac{2}{\left( \dfrac{2\tan ({{15}^{\circ }})}{1-{{\tan }^{2}}({{15}^{\circ }})} \right)}=\dfrac{2}{\tan \left( 2\times {{15}^{\circ }} \right)}=\dfrac{2}{\tan \left( {{30}^{\circ }} \right)}..................(1.5)$$ Now, the value of\tan \left( {{30}^{\circ }} \right)$is equal to$\dfrac{1}{\sqrt{3}}$. So, putting this value of$\tan \left( {{30}^{\circ }} \right)$in equation 1.5, we get:$\dfrac{2}{\tan \left( {{30}^{\circ }} \right)}=\dfrac{2}{\dfrac{1}{\sqrt{3}}}=2\sqrt{3}$Now, this value is equal to the Right Hand Side (RHS) of the equation in the question and also is derived from the Left Hand Side (LHS). Hence, we have satisfied the condition for proving, that is LHS=RHS. Therefore, we have proved that$\cot \left( {{105}^{\circ }} \right)-\tan \left( {{105}^{\circ }} \right)=2\sqrt{3}$.

Note: Here, the formula for the tangent of twice of an angle is derived from the formula for the tangent of the sum of two angles, that is:
$\tan \left( A+B \right)=\dfrac{\tan A+\tan B}{1-\tan A\tan B}$, by putting A in place of B as A+A=2A.For these types of problems always try to break the given angle into sum of angle angle and ${{90}^{\circ }}$ or ${{180}^{\circ }}$ for easier calculation.Students should remember important trigonometric formulas and identities for solving these types of questions.