JEE Mathematics 2026 Question with Solution

Given: We need to evaluate $E = \cosec 10^\circ - \sqrt{3} \sec 10^\circ$.

Find: The numerical value of the given trigonometric expression.

Convert cosec and sec into sine and cosine:

$$E = \frac{1}{\sin 10^\circ} - \frac{\sqrt{3}}{\cos 10^\circ}$$

Taking the common denominator,

$$E = \frac{\cos 10^\circ - \sqrt{3}\sin 10^\circ}{\sin 10^\circ \cos 10^\circ}$$

Now simplify the numerator:

$$\cos 10^\circ - \sqrt{3}\sin 10^\circ = 2\left(\frac{1}{2}\cos 10^\circ - \frac{\sqrt{3}}{2}\sin 10^\circ\right)$$

Using $\cos 60^\circ = \frac{1}{2}$ and $\sin 60^\circ = \frac{\sqrt{3}}{2}$,

$$\cos 10^\circ - \sqrt{3}\sin 10^\circ = 2(\cos 60^\circ \cos 10^\circ - \sin 60^\circ \sin 10^\circ)$$ $$= 2\cos(60^\circ + 10^\circ) = 2\cos 70^\circ$$

Now simplify the denominator using $\sin 2\theta = 2\sin \theta \cos \theta$:

$$\sin 10^\circ \cos 10^\circ = \frac{1}{2}\sin 20^\circ$$

Substituting back,

$$E = \frac{2\cos 70^\circ}{\frac{1}{2}\sin 20^\circ} = \frac{4\cos 70^\circ}{\sin 20^\circ}$$

Using the complementary angle identity $\cos 70^\circ = \sin 20^\circ$,

$$E = \frac{4\sin 20^\circ}{\sin 20^\circ} = 4$$

Therefore, the value of the expression is $4$. Hence, the correct option is D.

Given: $E = \cosec 10^\circ - \sqrt{3} \sec 10^\circ$.

Find: Its numerical value.

Write the expression over a common denominator:

$$E = \frac{\cos 10^\circ - \sqrt{3}\sin 10^\circ}{\sin 10^\circ \cos 10^\circ}$$

Now observe that the numerator matches the form of $\cos(A+B)$:

$$\cos 10^\circ - \sqrt{3}\sin 10^\circ = 2\cos 70^\circ$$

Also,

$$\sin 10^\circ \cos 10^\circ = \frac{1}{2}\sin 20^\circ$$

So,

$$E = \frac{2\cos 70^\circ}{\frac{1}{2}\sin 20^\circ} = \frac{4\cos 70^\circ}{\sin 20^\circ}$$

Since $\cos 70^\circ = \sin 20^\circ$,

$$E = 4$$

This works quickly because the numerator is intentionally convertible into a single cosine term. Therefore, the correct option is D.