Question

The value of $\tan {81}^0-\tan {63}^0-\tan {27}^0+\tan 9^0$is
$$\begin{gathered} \left(A\right).\text{ 0} \\ \left(B\right).\text{ 2} \\ \left(C\right).\text{ 3} \\ \left(D\right).\text{ 4} \end{gathered}$$

Answer 1

Hint: Try to write $\tan \theta$ in terms of $\cot \theta$, then use trigonometric formulas.

Given:$\tan {81}^0-\tan {63}^0-\tan {27}^0+\tan 9^0$
Rewriting above as:
$$\begin{gathered} \Rightarrow \left(\tan 9^0+\tan {81}^0\right)-\left(\tan {27}^0+\tan {63}^0\right) \\ \Rightarrow \left(\tan 9^0+\tan \left({90}^0-9^0\right)\right)-\left(\tan {27}^0+\tan \left({90}^0-{27}^0\right)\right)\dots \left(A\right) \end{gathered}$$
We know that, $$\tan \left({90}^0-\theta \right)=\cot \theta ={\displaystyle \frac{\cos \theta }{\sin \theta }}\dots \left(1\right)$$
Putting the value of $\tan \left({90}^0-\theta \right)$from $\left(1\right)$in $\left(A\right)$, we get
$\Rightarrow \tan 9^0+\cot 9^0-\tan {27}^0-\cot {27}^0$
Now, since $\tan \theta ={\displaystyle \frac{\sin \theta }{\cos \theta }}$ and $\cot \theta ={\displaystyle \frac{\cos \theta }{\sin \theta }}$
$$\begin{gathered} \therefore \left({\displaystyle \frac{\sin 9^0}{\cos 9^0}}+{\displaystyle \frac{\cos 9^0}{\sin 9^0}}\right)-\left({\displaystyle \frac{\sin {27}^0}{\cos {27}^0}}+{\displaystyle \frac{\cos {27}^0}{\sin {27}^0}}\right) \\ \Rightarrow \left({\displaystyle \frac{{\sin }^2{9}^0+{\cos }^2{9}^0}{\sin 9^0\cos 9^0}}\right)-\left({\displaystyle \frac{{\sin }^2{27}^0+{\cos }^2{27}^0}{\sin {27}^0\cos {27}^0}}\right) \end{gathered}$$
Using the property ${\sin }^2\theta +{\cos }^2\theta =1$ in above equation, we get
$$\Rightarrow \left({\displaystyle \frac{1}{\sin 9^0\cos 9^0}}\right)-\left({\displaystyle \frac{1}{\sin {27}^0\cos {27}^0}}\right)$$
Also, we know that $\sin x\cos x={\displaystyle \frac{\sin 2x}{2}}$
$$\begin{gathered} \therefore \left({\displaystyle \frac{2}{\sin {18}^0}}\right)-\left({\displaystyle \frac{2}{\sin {54}^0}}\right) \\ \Rightarrow 2\left({\displaystyle \frac{\sin {54}^0-\sin {18}^0}{\sin {18}^0\sin {54}^0}}\right) \end{gathered}$$
Now, by using $\sin A-\sin B=2\cos \left({\displaystyle \frac{A+B}{2}}\right)\sin \left({\displaystyle \frac{A-B}{2}}\right)$ in above equation, we get
$$\Rightarrow 2\left({\displaystyle \frac{2\cos \left({\displaystyle \frac{54+18}{2}}\right)\sin \left({\displaystyle \frac{54-18}{2}}\right)}{\sin {18}^0\sin \left({90}^0-{36}^0\right)}}\right)$$
using $\sin \left({90}^0-\theta \right)=\cos \theta$in the above equation, we get
$$\begin{gathered} \Rightarrow 2\left({\displaystyle \frac{2\cos \left({\displaystyle \frac{72}{2}}\right)\sin \left({\displaystyle \frac{36}{2}}\right)}{\sin {18}^0\cos {36}^0}}\right) \\ \Rightarrow 2\left({\displaystyle \frac{2\cos \left({36}^0\right)\sin \left({18}^0\right)}{\sin {18}^0\cos {36}^0}}\right) \\ \Rightarrow 4\left({\displaystyle \frac{\sin {18}^0\cos {36}^0}{\sin {18}^0\cos {36}^0}}\right) \\ \Rightarrow 4 \end{gathered}$$
$\therefore$the correct option is $\left(D\right)$.

Note: Whenever there are integer angles inside trigonometric functions whose values are not known to us, always try to convert them by using ${90}^0-\theta$ in order to make calculation easier.