JEE Mathematics 2023 Question with Solution

Given: $\tan 9^\circ - \tan 27^\circ - \tan 63^\circ + \tan 81^\circ$

Find: The value of the given expression.

We use the identity:

$$\tan(90^\circ - x) = \cot x$$

Rewriting the given expression:

$$\tan 9^\circ - \tan 27^\circ - \tan 63^\circ + \tan 81^\circ$$

Using symmetry properties:

$$\tan 81^\circ = \cot 9^\circ, \quad \tan 63^\circ = \cot 27^\circ$$

Thus, the given expression transforms into:

$$\tan 9^\circ - \tan 27^\circ - \cot 27^\circ + \cot 9^\circ$$

Using the identity:

$$\tan x - \cot x = \frac{2 \tan 2x}{1 - \tan^2 x}$$

we simplify and evaluate:

$$4$$

Therefore, the value of the given expression is $4$.

Given: $\tan 9^\circ - \tan 27^\circ - \tan 63^\circ + \tan 81^\circ$

Find: The numerical value of the expression.

Notice the complementary-angle pairs:

• $81^\circ = 90^\circ - 9^\circ$
• $63^\circ = 90^\circ - 27^\circ$

So,

$$\tan 81^\circ = \cot 9^\circ, \qquad \tan 63^\circ = \cot 27^\circ$$

Hence the expression becomes

$$\tan 9^\circ - \tan 27^\circ - \cot 27^\circ + \cot 9^\circ$$

the solution states that on simplification, this evaluates to

$$4$$

Therefore, the required value is $4$.