Question

$\tan (81°)-\tan (63°)-\tan (27°)+\tan (9°)$ equals

• A. $6$
• B. $0$
• C. $2$
• D. $4$

Answer 1

The correct option is D

$4$

Explanation for correct option:

Evaluating the given expression:

$\tan (81°)-\tan (63°)-\tan (27°)+\tan (9°)=\tan (90°-9)+\tan (9°)-\tan (90°-27°)-\tan (27°)$ $$\begin{gathered} =\cot \left(9°\right)+\tan (9°)-\cot \left(27°\right)-\tan (27°)\left[\because \tan (90-\theta )=\cot \theta \right] \\ =\frac{\cos (9°)}{\sin (9°)}+\frac{\sin (9°)}{\cos (9°)}-\left(\frac{\cos (27°)}{\sin (27°)}+\frac{\sin (27°)}{\cos (27°)}\right)\left[\because \cot \theta =\frac{\cos \theta }{\sin \theta }\right] \\ =\frac{{\cos }^2(9°)+{\sin }^2\left(9°\right)}{\sin (9°)\cos (9°)}-\left(\frac{{\cos }^2(27°)+{\sin }^2(27°)}{\sin (27°)\cos (27°)}\right) \\ =\frac{2}{\sin (18°)}-\frac{2}{\sin (54°)}\mathbf{\left[}\mathbf{\because }\mathbf{}\mathbf{sin}\mathbf{\left(}\mathbf{2}\mathbf{x}\mathbf{\right)}\mathbf{=}\mathbf{2}\mathbf{sinx}\mathbf{}\mathbf{cosx}\mathbf{,}\mathbf{}{\mathbf{cos}}^{\mathbf{2}}\mathbf{\right(}\mathbf{\theta }\mathbf{)}\mathbf{+}{\mathbf{sin}}^{\mathbf{2}}\mathbf{(}\mathbf{\theta }\mathbf{)}\mathbf{=}\mathbf{1}\mathbf{]} \end{gathered}$$

$$\begin{gathered} =\frac{2\left(\sin (54°)-\sin (18°)\right)}{\sin (18°)\sin (54°)} \\ =\frac{2\left(2\cos \right(36°\left)\sin \right(18°\left)\right)}{\cos (36°)\sin (18°)}\mathbf{}\mathbf{[}\mathbf{\because }\mathbf{}\mathbf{sinx}\mathbf{-}\mathbf{siny}\mathbf{=}\mathbf{2}\mathbf{cos}\mathbf{\left(}\frac{\mathbf{a}\mathbf{+}\mathbf{b}}{\mathbf{2}}\mathbf{\right)}\mathbf{sin}\mathbf{\left(}\frac{\mathbf{a}\mathbf{-}\mathbf{b}}{\mathbf{2}}\mathbf{\right)}\mathbf{]} \\ =4 \end{gathered}$$

Hence, the correct answer is option (D).