Question

If \[\cos A=\dfrac{\sqrt{3}}{2}\], then $\tan 3A=$

Answer 1

Hint: Now we are given that $\cos A=\dfrac{\sqrt{3}}{2}$ . Now we will use the identity ${{\sin }^{2}}x=1-{{\cos }^{2}}x$ to find the value of ${{\sin }^{2}}A$ . Hence taking square root in the equation obtained we will find the value of $\sin A$ . Now we know that $\tan x=\dfrac{\sin x}{\cos x}$ . Hence using this we will find the value of $\tan A$ . Now we will substitute the value of $\tan A$ in the formula \[\tan 3A=\dfrac{3\tan A-{{\tan }^{3}}A}{1-3{{\tan }^{2}}A}\] and hence find the value of $\tan 3A$ .

Complete step by step solution:
Now we are given that the value of $\cos A$ is $\dfrac{\sqrt{3}}{2}$ .
Now first we will find the value of $\sin A$ .
Consider the equation $\cos A=\dfrac{\sqrt{3}}{2}$
Now squaring the above equation we get, ${{\cos }^{2}}A=\dfrac{3}{4}$ .
Now let us multiply the above equation by – 1 Hence we get,
$\Rightarrow -{{\cos }^{2}}A=\dfrac{3}{4}$
Now Adding 1 to both sides of the equation we get,
$\Rightarrow 1-{{\cos }^{2}}A=1-\dfrac{3}{4}$
Now we know that ${{\sin }^{2}}x=1-{{\cos }^{2}}x$ Hence using this we get,
$\begin{align}
  & \Rightarrow {{\sin }^{2}}x=\dfrac{4-3}{4} \\
 & \Rightarrow {{\sin }^{2}}x=\dfrac{1}{4} \\
\end{align}$
Now taking square root on both sides we get,
$\Rightarrow \sin A=\dfrac{1}{2}$
Now we have the value of $\sin A$ and $\cos A$ . Hence we can easily find the value of $\tan A$.
Hence we get,
$\Rightarrow \tan A=\dfrac{\sin A}{\cos A}=\dfrac{\dfrac{1}{2}}{\dfrac{\sqrt{3}}{2}}=\dfrac{1}{\sqrt{3}}$
Hence we have $\tan A=\dfrac{1}{\sqrt{3}}$ .
Now consider $\tan 3A$ .
We know by compound angles formula that $\tan 3\theta =\dfrac{3\tan \theta -{{\tan }^{3}}\theta }{1-3{{\tan }^{2}}\theta }$
Now substituting the value of $\tan A$ in the formula we get,
$\Rightarrow \tan 3A=\dfrac{3\times \dfrac{1}{\sqrt{3}}-{{\left( \dfrac{1}{\sqrt{3}} \right)}^{3}}}{1-3{{\left( \dfrac{1}{\sqrt{3}} \right)}^{2}}}$
Now on simplifying we can see that the denominator is coming to be 0.

Hence the value of $\tan 3A$ is nothing but $\infty $

Note: Now note that we can directly solve the given problem by finding the value of A. Here we are given that $\cos A=\dfrac{\sqrt{3}}{2}$ . taking the inverse of $\cos $ on both side we get $A=\dfrac{\pi }{6}$ . Now since $A=\dfrac{\pi }{6}$ we can say that the value of $3A=\dfrac{\pi }{2}$ . Now we know that $\tan \left( \dfrac{\pi }{2} \right)=\infty $