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

Find the value of \(\frac{2}{3}\)tan2 60° + 3 cos2 30° − 2 sec2 30° − \(\frac{3}{4}\)cot2 60°.

This question was previously asked in
SSC CGL 2023 (Tier-II) Paper 1 Previous Year Paper (26-Oct-2023) (Shift-1)
The correct answer is

1.33

Evaluating Trigonometric Expression

The problem asks us to find the value of a given trigonometric expression involving standard angles 30° and 60°. The expression is:

\(\frac{2}{3}\)tan² 60° + 3 cos² 30° − 2 sec² 30° − \(\frac{3}{4}\)cot² 60°

To evaluate this expression, we need to know the values of the trigonometric ratios for 30° and 60°.

Key Trigonometric Values for 30° and 60°

Here are the required standard trigonometric values:

Ratio 30° 60°
tan \(\frac{1}{\sqrt{3}}\) \(\sqrt{3}\)
cos \(\frac{\sqrt{3}}{2}\) \(\frac{1}{2}\)
sec \(\frac{2}{\sqrt{3}}\) 2
cot \(\sqrt{3}\) \(\frac{1}{\sqrt{3}}\)

Now, let's substitute these values into the given expression.

Step-by-Step Calculation

The expression is \(\frac{2}{3}\)tan² 60° + 3 cos² 30° − 2 sec² 30° − \(\frac{3}{4}\)cot² 60°.

Substitute the values:

  • tan 60° = \(\sqrt{3}\), so tan² 60° = \((\sqrt{3})^2 = 3\)
  • cos 30° = \(\frac{\sqrt{3}}{2}\), so cos² 30° = \((\frac{\sqrt{3}}{2})^2 = \frac{3}{4}\)
  • sec 30° = \(\frac{2}{\sqrt{3}}\), so sec² 30° = \((\frac{2}{\sqrt{3}})^2 = \frac{4}{3}\)
  • cot 60° = \(\frac{1}{\sqrt{3}}\), so cot² 60° = \((\frac{1}{\sqrt{3}})^2 = \frac{1}{3}\)

Substitute these squared values back into the expression:

\(\frac{2}{3}(3) + 3(\frac{3}{4}) - 2(\frac{4}{3}) - \frac{3}{4}(\frac{1}{3})\)

Now, simplify each term:

  • \(\frac{2}{3}(3) = 2\)
  • \(3(\frac{3}{4}) = \frac{9}{4}\)
  • \(2(\frac{4}{3}) = \frac{8}{3}\)
  • \(\frac{3}{4}(\frac{1}{3}) = \frac{3}{12} = \frac{1}{4}\)

Substitute the simplified terms back into the expression:

\(2 + \frac{9}{4} - \frac{8}{3} - \frac{1}{4}\)

Group the terms with the same denominators:

\(2 + (\frac{9}{4} - \frac{1}{4}) - \frac{8}{3}\)

Calculate the terms in the parenthesis:

\(\frac{9}{4} - \frac{1}{4} = \frac{9-1}{4} = \frac{8}{4} = 2\)

Substitute this back:

\(2 + 2 - \frac{8}{3}\)

\(4 - \frac{8}{3}\)

To subtract the fraction from the whole number, find a common denominator, which is 3:

\(\frac{4 \times 3}{3} - \frac{8}{3}\)

\(\frac{12}{3} - \frac{8}{3}\)

\(\frac{12 - 8}{3}\)

\(\frac{4}{3}\)

Convert the fraction to a decimal:

\(\frac{4}{3} \approx 1.333...\)

Rounding to two decimal places, the value is approximately 1.33.

Revision Table: Standard Angle Trigonometric Values

Angle \(\theta\) sin \(\theta\) cos \(\theta\) tan \(\theta\) csc \(\theta\) sec \(\theta\) cot \(\theta\)
0° (0 rad) 0 1 0 Undefined 1 Undefined
30° (\(\frac{\pi}{6}\) rad) \(\frac{1}{2}\) \(\frac{\sqrt{3}}{2}\) \(\frac{1}{\sqrt{3}}\) 2 \(\frac{2}{\sqrt{3}}\) \(\sqrt{3}\)
45° (\(\frac{\pi}{4}\) rad) \(\frac{1}{\sqrt{2}}\) \(\frac{1}{\sqrt{2}}\) 1 \(\sqrt{2}\) \(\sqrt{2}\) 1
60° (\(\frac{\pi}{3}\) rad) \(\frac{\sqrt{3}}{2}\) \(\frac{1}{2}\) \(\sqrt{3}\) \(\frac{2}{\sqrt{3}}\) 2 \(\frac{1}{\sqrt{3}}\)
90° (\(\frac{\pi}{2}\) rad) 1 0 Undefined 1 Undefined 0

Additional Information: Trigonometric Identities

This problem uses trigonometric ratios at standard angles. It's also important to remember the reciprocal identities that relate some trigonometric functions:

  • sec \(\theta = \frac{1}{\text{cos } \theta}\)
  • csc \(\theta = \frac{1}{\text{sin } \theta}\)
  • cot \(\theta = \frac{1}{\text{tan } \theta}\) or cot \(\theta = \frac{\text{cos } \theta}{\text{sin } \theta}\)

Also, the square notation like tan² \(\theta\) means \((\text{tan } \theta)^2\), cos² \(\theta\) means \((\text{cos } \theta)^2\), and so on.

Being familiar with these identities and the standard angle values is crucial for solving trigonometric expressions and equations.

Was this answer helpful?

Similar Questions

  1. Simplify the following.

    \(\frac{\sin^3 α + \cos^3 α}{\sin α + \cos α}\)

  2. Find the value of the following expression.

    12(sin4 θ + cos4 θ) + 18(sinθ + cos6 θ) + 78 sin2 θ cos2 θ

  3. Find the exact value of cos 120°.

  4. Simplify \(\frac{1+sint}{4-4sint}-\frac{1-sint}{4+4sint}\)

  5. Find the value of tan (−1125°).

  6. ΔABC is a right triangle. If ∠B = 90° and tan A = \(\frac{1}{\sqrt2}\), then the value of sin A cos C + cos A sin C is :

  7. If \(\rm \frac{21\ cosA+3\ sinA}{3\ cosA+4\ sinA}\) = 2, then find the value of cot A

  8. In a right triangle for an acute angle x, if sin x = \(\frac{3}{7}\), then find the value of cosx.

  9. If \(\rm cos x + sec x = {7 \over2 \sqrt3}\), then the value of cos2x + sec2x will be ________.

  10. If \(\rm tan A = {3 \over 8},\) then the value of \({3 \sin A + 2 \cos A} \over 3 \sin A - 2 \cos A\) is:


Important Questions from Trigonometry

  1. The value of 5 sin 14° sec 76° + 3 cot 15° cot 75° + 2 tan 45° is:

  2. If two complimentary angles are in the ratio of 4 : 5, find the greater angle.

  3. If \(\frac{\sin\spaceθ \space+\space \cos\spaceθ} {\sin \spaceθ \space-\space \cos \spaceθ} = \frac{\sqrt3 \space-\space 1}{\sqrt3 \space+\space 1} \) , then the angle θ is 

  4. If tan α = 1/2, tan β = 1/3, then find α + β.

  5. Simplify: sin (A + B) sin (A – B)

Need Expert Advice?
Upcoming Exams
SSC JHT
September 08, 2026
SSC Stenographer
September 09, 2026
SSC Selection Post
September 16, 2026
Test Series
SSC CGL img
SSC
SSC CGL (Tier I + Tier II) 2026 Mock Test Series - Latest Pattern
2500 Tests 6 Tests Free
3973 Attempts
4.2(838)
English, Hindi

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