Question

# Find the value of $\tan 9^\circ - \tan 27^\circ - \tan 63^\circ + \tan 81^\circ$.A.-1B.0C.1D.4

Hint: We need to know the formulae of trigonometric functions in different quadrants and basic values of trigonometric functions to solve the given problem.

Given expression is $\tan 9^\circ - \tan 27^\circ - \tan 63^\circ + \tan 81^\circ$
$\left[ {\because \tan \theta = \cot (90 - \theta )} \right]$, So we can write
$= \tan 9^\circ + \cot 9^\circ - \left( {\tan 27^\circ + \cot 27^\circ } \right)$
$= \frac{{1 + {{\tan }^2}9^\circ }}{{\tan 9^\circ }} - \frac{{1 + {{\tan }^2}27^\circ }}{{\tan 27^\circ }}$ (+1 and -1 get cancelled out)
$\left[ {\because 1 + {{\tan }^2}\theta = {{\sec }^2}\theta } \right]$
$= \frac{{{{\sec }^2}9^\circ }}{{\tan 9^\circ }} - \frac{{{{\sec }^2}27^\circ }}{{\tan 27^\circ }}$
We can simply the above expression by writing as
$= \frac{1}{{\sin 9^\circ \cos 9^\circ }} - \frac{1}{{\sin 27^\circ \cos 27^\circ }}$ $\left[ {\because \sec \theta = \frac{1}{{\cos \theta }}\& \tan \theta = \frac{{\sin \theta }}{{\cos \theta }}} \right]$
Multiplying and dividing the above term with 2
$= \frac{2}{{\sin 18^\circ }} - \frac{2}{{\sin 54^\circ }}$
$= \frac{2}{{\frac{{\sqrt 5 - 1}}{4}}} - \frac{2}{{\frac{{\sqrt 5 + 1}}{4}}}$
$= 8\left( {\frac{1}{{\sqrt 5 - 1}} - \frac{1}{{\sqrt 5 + 1}}} \right)$
$= 8\left( {\frac{{\sqrt 5 + 1 - \sqrt 5 + 1}}{4}} \right)$
$= 8\left( {\frac{2}{4}} \right) = 2 \times 2 = 4$
$\therefore$ The value of $\tan 9^\circ - \tan 27^\circ - \tan 63^\circ + \tan 81^\circ$= 4
Note: $81^\circ$ and $63^\circ$ lies in the first quadrant. Here if we observe $81^\circ$ and $9^\circ$ are complementary angles. Similarly $63^\circ$ and $27^\circ$ are complementary angles. Using this idea, we simplified them into a single trigonometric function. The value of $\sin 18^\circ = \frac{{\sqrt 5 - 1}}{4}$ and the value of$\sin 54^\circ = \frac{{\sqrt 5 + 1}}{4}$.