The value of cosec 10° - √3 sec 10° is equal to :

Q. The value of \(\text{cosec } 10^\circ - \sqrt{3} \sec 10^\circ\) is equal to :

(A) 2

(B) 6

(C) 8

(D) 4

Correct Answer: (D) 4

The final simplified trigonometric value is 4

Step-by-Step Detailed Solution

Step 1: Convert to Sine and Cosine

The given expression involves cosecant and secant. To simplify, we first rewrite them in terms of their reciprocal functions, sine and cosine:

Expression = \(\frac{1}{\sin 10^\circ} - \frac{\sqrt{3}}{\cos 10^\circ}\)

Step 2: Take the Common Denominator

Combine the fractions by finding a common denominator, which is \(\sin 10^\circ \cos 10^\circ\):

Expression = \(\frac{\cos 10^\circ - \sqrt{3} \sin 10^\circ}{\sin 10^\circ \cos 10^\circ}\)

Step 3: Transform the Numerator

We use the method of "Multiplying and Dividing by 2" to convert the numerator into a single trigonometric ratio. Since \(\sqrt{1^2 + (\sqrt{3})^2} = 2\), we factor out a 2:

Numerator = \(2 \left( \frac{1}{2} \cos 10^\circ - \frac{\sqrt{3}}{2} \sin 10^\circ \right)\) Now substitute \(\frac{1}{2} = \sin 30^\circ\) and \(\frac{\sqrt{3}}{2} = \cos 30^\circ\): Numerator = \(2 (\sin 30^\circ \cos 10^\circ - \cos 30^\circ \sin 10^\circ)\)

Step 4: Apply the Sine Difference Identity

Recall the identity \(\sin(A - B) = \sin A \cos B - \cos A \sin B\). In our case, \(A = 30^\circ\) and \(B = 10^\circ\):

Numerator = \(2 \sin(30^\circ - 10^\circ) = 2 \sin 20^\circ\)

Step 5: Transform the Denominator

Use the double angle formula \(\sin 2\theta = 2 \sin \theta \cos \theta\), which implies \(\sin \theta \cos \theta = \frac{1}{2} \sin 2\theta\):

Denominator = \(\sin 10^\circ \cos 10^\circ = \frac{1}{2} \sin(2 \times 10^\circ) = \frac{1}{2} \sin 20^\circ\)

Step 6: Final Calculation

Substitute the simplified numerator and denominator back into the main expression:

Expression = \(\frac{2 \sin 20^\circ}{\frac{1}{2} \sin 20^\circ}\) Expression = \(2 \times 2 \times \frac{\sin 20^\circ}{\sin 20^\circ}\) Expression = 4

Related Theory: Trigonometric Transformations

Trigonometric expressions involving uncommon angles like 10°, 20°, or 40° are a staple in JEE Main. The core strategy always involves reducing them to standard angles (30°, 45°, 60°) or creating situations where ratios cancel out using identities.

1. The \(a \cos \theta + b \sin \theta\) Form

A very common technique in competitive exams is simplifying expressions of the form \(a \sin \theta \pm b \cos \theta\). To do this, we multiply and divide by \(\sqrt{a^2 + b^2}\). This allows the expression to be converted into either a single sine or cosine wave:

\(a \cos \theta - b \sin \theta = \sqrt{a^2 + b^2} \left( \frac{a}{\sqrt{a^2+b^2}} \cos \theta - \frac{b}{\sqrt{a^2+b^2}} \sin \theta \right)\) Let \(\sin \phi = \frac{a}{\sqrt{a^2+b^2}}\) and \(\cos \phi = \frac{b}{\sqrt{a^2+b^2}}\). Then it becomes \(\sqrt{a^2+b^2} \sin(\phi - \theta)\).

2. Compound Angle Identities

These are the backbone of most trigonometry problems in JEE:

\(\sin(A \pm B) = \sin A \cos B \pm \cos A \sin B\)
\(\cos(A \pm B) = \cos A \cos B \mp \sin A \sin B\)
\(\tan(A \pm B) = \frac{\tan A \pm \tan B}{1 \mp \tan A \tan B}\)

3. Double and Triple Angle Formulas

When you see terms like \(\sin \theta \cos \theta\), immediately think of \(\sin 2\theta\):

\sin 2\theta = 2 \sin \theta \cos \theta\) \(\cos 2\theta = \cos^2 \theta - \sin^2 \theta = 2\cos^2 \theta - 1 = 1 - 2\sin^2 \theta

Triple angle formulas are equally important for harder shifts:

\sin 3\theta = 3 \sin \theta - 4 \sin^3 \theta\) \(\cos 3\theta = 4 \cos^3 \theta - 3 \cos \theta

4. Product-to-Sum and Sum-to-Product

Useful for simplifying series of trigonometric terms:

2 \sin A \cos B = \sin(A+B) + \sin(A-B)\) \(\sin C + \sin D = 2 \sin(\frac{C+D}{2}) \cos(\frac{C-D}{2})

5. Values of Standard Non-Standard Angles

While 10° is solved through identities, some values are worth memorizing to save time:

\(\sin 18^\circ = \frac{\sqrt{5}-1}{4}\)
\(\cos 36^\circ = \frac{\sqrt{5}+1}{4}\)
\(\sin 15^\circ = \frac{\sqrt{3}-1}{2\sqrt{2}}\)

6. The Shortcut Approach for JEE

In many MCQ scenarios, if you encounter an expression like \(\text{cosec } \theta - \sqrt{3} \sec \theta\), the result is often \(4 \frac{\sin(60 - \theta)}{\sin 2\theta}\). For \(\theta = 10\), this gives \(4 \frac{\sin 50}{\sin 20}\), but the identity we used in the step-by-step is more direct for specific coefficients like \(\sqrt{3}\).

7. Analyzing the Coefficient \(\sqrt{3}\)

Whenever you see \(\sqrt{3}\) paired with trigonometric terms, it is almost always an invitation to use \(\tan 60^\circ\), \(\sqrt{3}/2\) (which is \(\sin 60^\circ\) or \(\cos 30^\circ\)), or \(1/\sqrt{3}\). Identifying this early is key to choosing the right substitution.

8. Common Mistake: Reciprocal Confusion

Students often confuse \(\text{cosec}\) with \(\cos\) and \(\sec\) with \(\sin\). Always double-check: \(\text{cosec } x = 1/\sin x\) and \(\sec x = 1/\cos x\). Swapping these results in an entirely different (and usually unsolvable) expression.

9. Strategy for Similar Questions

If you see \(\text{cosec } 20^\circ - \dots\), or complex fractions, always reduce everything to \(\sin\) and \(\cos\) first. This provides a clear path for cross-multiplication and identity application.

10. Exam Importance

This specific question template (\(\text{cosec } A - \sqrt{3} \sec A\)) has appeared multiple times in JEE Main over the last decade. It tests the student's ability to recognize the \(a \sin \theta + b \cos \theta\) form under a layer of reciprocal functions.

Frequently Asked Questions (FAQs)

1. Why did we multiply and divide the numerator by 2? To normalize the coefficients (1 and √3) so that they become standard trigonometric values like 1/2 and √3/2, which are the sine and cosine of 30° and 60°.

2. Could we use cos 60° instead of sin 30°? Yes. You could write it as 2(cos 60° cos 10° - sin 60° sin 10°), which simplifies to 2 cos(60° + 10°) = 2 cos 70°. Since cos 70° = sin 20°, the result remains the same.

3. What is the fundamental identity used in the denominator? The double angle identity: sin 2θ = 2 sin θ cos θ. By rearranging, sin θ cos θ = (sin 2θ)/2.

4. Is the value of sin 20°/sin 20° always 1? Yes, as long as the angle is not an integer multiple of 180°, sin θ is non-zero, allowing us to cancel it.

5. What happens if the expression was cosec 10° + √3 sec 10°? The numerator would become 2 sin(30° + 10°) = 2 sin 40°. The denominator would still be (1/2) sin 20°. The result would then involve further simplification using sin 40° = 2 sin 20° cos 20°.

6. How can I identify that I need to convert to sin and cos? In JEE, 90% of problems involving sec, cosec, tan, or cot are easier to solve once reduced to their basic sin and cos components.

7. Why is √3 significant here? √3 is the tangent of 60° or twice the cosine of 30°. Its presence usually signals the use of 30-60-90 triangle ratios.

8. Is this a common board exam question too? Yes, this is a standard problem in Class 11 Trigonometry and is frequently asked in competitive engineering exams.

9. What is the value of cosec 10° roughly? Since sin 10° is small (~0.1736), cosec 10° is quite large, approximately 5.76.

10. Can I use a calculator in JEE Main? No, calculators are not allowed. You must rely on trigonometric identities and algebraic manipulation.

Expert Contribution by: Roshan

JEE NEET Expert | 10+ Years Experience | Specializing in Trigonometry and Calculus.

The value of cosec 10° – √3 sec 10° is equal to :