All Exams Test series for 1 year @ ₹349 only
Question

The value of \(\frac{tan^2 30^\circ+sin^290^\circ+cot^260^\circ+sin^230^\circ cos^245^\circ}{sin60^\circ cos30^\circ-cos60^\circ sin30^\circ}\)

This question was previously asked in
SSC CGL 2020 Tier-II (English) Previous Year Paper (29-Jan-2022)
The correct answer is

43/12

Evaluating Trigonometric Expression with Standard Angles

The question asks us to find the value of a given trigonometric expression:

\[ \frac{\tan^2 30^\circ + \sin^2 90^\circ + \cot^2 60^\circ + \sin^2 30^\circ \cos^2 45^\circ}{\sin 60^\circ \cos 30^\circ - \cos 60^\circ \sin 30^\circ} \]

To evaluate this expression, we need to know the values of the trigonometric functions for the standard angles \(30^\circ, 45^\circ, 60^\circ,\) and \(90^\circ\).

Standard Trigonometric Values

Here are the standard values we will use:

Angle \(\sin \theta\) \(\cos \theta\) \(\tan \theta\) \(\cot \theta\)
\(30^\circ\) \(\frac{1}{2}\) \(\frac{\sqrt{3}}{2}\) \(\frac{1}{\sqrt{3}}\) \(\sqrt{3}\)
\(45^\circ\) \(\frac{1}{\sqrt{2}}\) \(\frac{1}{\sqrt{2}}\) \(1\) \(1\)
\(60^\circ\) \(\frac{\sqrt{3}}{2}\) \(\frac{1}{2}\) \(\sqrt{3}\) \(\frac{1}{\sqrt{3}}\)
\(90^\circ\) \(1\) \(0\) Undefined \(0\)

Evaluate the Numerator

The numerator is \(\tan^2 30^\circ + \sin^2 90^\circ + \cot^2 60^\circ + \sin^2 30^\circ \cos^2 45^\circ\).

Substitute the standard values:

  • \(\tan 30^\circ = \frac{1}{\sqrt{3}}\), so \(\tan^2 30^\circ = \left(\frac{1}{\sqrt{3}}\right)^2 = \frac{1}{3}\).
  • \(\sin 90^\circ = 1\), so \(\sin^2 90^\circ = (1)^2 = 1\).
  • \(\cot 60^\circ = \frac{1}{\sqrt{3}}\), so \(\cot^2 60^\circ = \left(\frac{1}{\sqrt{3}}\right)^2 = \frac{1}{3}\).
  • \(\sin 30^\circ = \frac{1}{2}\) and \(\cos 45^\circ = \frac{1}{\sqrt{2}}\), so \(\sin^2 30^\circ \cos^2 45^\circ = \left(\frac{1}{2}\right)^2 \times \left(\frac{1}{\sqrt{2}}\right)^2 = \frac{1}{4} \times \frac{1}{2} = \frac{1}{8}\).

Now, add these values together for the numerator:

\[ \text{Numerator} = \frac{1}{3} + 1 + \frac{1}{3} + \frac{1}{8} \]

Combine the fractions with like denominators:

\[ \text{Numerator} = \left(\frac{1}{3} + \frac{1}{3}\right) + 1 + \frac{1}{8} = \frac{2}{3} + 1 + \frac{1}{8} \]

To add these, find a common denominator, which is 24:

\[ \text{Numerator} = \frac{2 \times 8}{3 \times 8} + \frac{1 \times 24}{1 \times 24} + \frac{1 \times 3}{8 \times 3} = \frac{16}{24} + \frac{24}{24} + \frac{3}{24} \]

\[ \text{Numerator} = \frac{16 + 24 + 3}{24} = \frac{43}{24} \]

Evaluate the Denominator

The denominator is \(\sin 60^\circ \cos 30^\circ - \cos 60^\circ \sin 30^\circ\).

Substitute the standard values:

  • \(\sin 60^\circ = \frac{\sqrt{3}}{2}\)
  • \(\cos 30^\circ = \frac{\sqrt{3}}{2}\)
  • \(\cos 60^\circ = \frac{1}{2}\)
  • \(\sin 30^\circ = \frac{1}{2}\)

So, the denominator is:

\[ \text{Denominator} = \left(\frac{\sqrt{3}}{2}\right) \left(\frac{\sqrt{3}}{2}\right) - \left(\frac{1}{2}\right) \left(\frac{1}{2}\right) \]

\[ \text{Denominator} = \frac{\sqrt{3} \times \sqrt{3}}{2 \times 2} - \frac{1 \times 1}{2 \times 2} = \frac{3}{4} - \frac{1}{4} \]

\[ \text{Denominator} = \frac{3 - 1}{4} = \frac{2}{4} = \frac{1}{2} \]

Alternatively, we can recognise the trigonometric identity \(\sin(A - B) = \sin A \cos B - \cos A \sin B\). The denominator has the form \(\sin 60^\circ \cos 30^\circ - \cos 60^\circ \sin 30^\circ\), which is \(\sin(60^\circ - 30^\circ) = \sin 30^\circ\). The value of \(\sin 30^\circ\) is \(\frac{1}{2}\). Both methods give the same result for the denominator.

Calculate the Final Value of the Expression

The expression is \(\frac{\text{Numerator}}{\text{Denominator}}\).

Substitute the calculated values:

\[ \text{Value} = \frac{\frac{43}{24}}{\frac{1}{2}} \]

To divide by a fraction, multiply by its reciprocal:

\[ \text{Value} = \frac{43}{24} \times \frac{2}{1} = \frac{43 \times 2}{24 \times 1} = \frac{86}{24} \]

Simplify the fraction by dividing both the numerator and the denominator by their greatest common divisor, which is 2:

\[ \text{Value} = \frac{86 \div 2}{24 \div 2} = \frac{43}{12} \]

Thus, the value of the given trigonometric expression is \(\frac{43}{12}\).

Revision Table: Key Trigonometric Values

Function \(30^\circ\) \(45^\circ\) \(60^\circ\) \(90^\circ\)
\(\sin\) \(\frac{1}{2}\) \(\frac{1}{\sqrt{2}}\) \(\frac{\sqrt{3}}{2}\) \(1\)
\(\cos\) \(\frac{\sqrt{3}}{2}\) \(\frac{1}{\sqrt{2}}\) \(\frac{1}{2}\) \(0\)
\(\tan\) \(\frac{1}{\sqrt{3}}\) \(1\) \(\sqrt{3}\) Undefined
\(\cot\) \(\sqrt{3}\) \(1\) \(\frac{1}{\sqrt{3}}\) \(0\)

Additional Information: Trigonometric Identities Used

This problem utilised the following concepts:

  • Values of trigonometric functions for standard angles (\(30^\circ, 45^\circ, 60^\circ, 90^\circ\)).
  • Properties of exponents with trigonometric functions, e.g., \(\sin^2 \theta = (\sin \theta)^2\).
  • Arithmetic operations with fractions.
  • Implicitly, the angle subtraction formula for sine: \(\sin(A - B) = \sin A \cos B - \cos A \sin B\). While the denominator calculation can be done by substituting values directly, recognizing this identity provides an alternative method and reinforces understanding of trigonometric formulas.
Was this answer helpful?

Similar Questions

  1. If \(sin\left(\frac{2A+B}{2}\right) = cos\left(\frac{2A-B}{2}\right)=\frac{\sqrt{3}}{2},0^{\circ}<\frac{2A+B}{2}<90^{\circ}\)  then find the value of sin[3(A - B)].

  2. Let x = r cos(t), y = r sin(t) cos(u), z = r sin(t) sin(u). Then the value of x² + y² + z² is:

     

  3. Find the value of \(\frac{\cos41}{\sin49} + \frac{\sin51}{\cos39}\)

  4. The value of θ, when √3 cos θ + sin θ = 1 (1 ≤ θ ≤ 90°), is:

  5. If 1/(1 - sin θ) + 1/(1 + sin θ) = 4 sec θ, (0 < θ < 90°), then the value of (cot θ + cosec θ) is:

  6. If sin θ - cos θ = \(4 \over5\), then find the value of sin θ + cos θ.


Important Questions from Trigonometric Functions

  1. If \(\tan A=\dfrac{1}{7}\) , then what is cos 2A equal to?

  2. What is tan 2A equal to?

  3. What is the value of tan 18°?

  4. What is the period of the function f(x) = ln(2 + sin 2x)?

  5. If tan (45° + θ) = 1 + sin 2θ, where \(-\frac{\pi}{4}< θ <\frac{\pi}{4},\)  then what is the value of cos 2θ?

Need Expert Advice?
Upcoming Exams
SSC CPO
November 23, 2026
SSC CHSL
November 30, 2026
Test Series
SSC CGL img
SSC
SSC CGL (Tier I + Tier II) 2026 Mock Test Series - Latest Pattern
2509 Tests 6 Tests Free
8926 Attempts
4.2(945)
English, Hindi
More Questions from SSC CGL

Start Your Preparation with Prepp Mobile App

Download the app from Google Play & App Store
Download the app from Google Play & App Store
Prepp Mobile App