NCERT Solutions Class 11 Maths Chapter 3 Exercise 3.2

Quick Concept Recap

This exercise deals with finding the values of trigonometric functions based on their signs in different quadrants.

- Quadrant I: All trigonometric functions are positive.

- Quadrant II: Only \(\sin x\) and \(\text{cosec } x\) are positive.

- Quadrant III: Only \(\tan x\) and \(\cot x\) are positive.

- Quadrant IV: Only \(\cos x\) and \(\sec x\) are positive.

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Detailed Solutions

Question 1: Find the values of other five trigonometric functions if \(\cos x = -\frac{1}{2}\), \(x\) lies in the third quadrant.

Solution:

Given: \(\cos x = -\frac{1}{2}\).

Since \(x\) is in the 3rd quadrant, only \(\tan x\) and \(\cot x\) are positive; others are negative.

1. Find \(\sec x\):

\[ \sec x = \frac{1}{\cos x} = \frac{1}{-1/2} = \mathbf{-2} \]

2. Find \(\sin x\) using \(\sin^2 x + \cos^2 x = 1\):

\[ \sin^2 x = 1 - \cos^2 x = 1 - \left(-\frac{1}{2}\right)^2 = 1 - \frac{1}{4} = \frac{3}{4} \]

\[ \sin x = \pm \frac{\sqrt{3}}{2} \]

Since \(x\) is in the 3rd quadrant, \(\sin x\) is negative:

\[ \sin x = \mathbf{-\frac{\sqrt{3}}{2}} \]

3. Find \(\text{cosec } x\):

\[ \text{cosec } x = \frac{1}{\sin x} = \frac{1}{-\sqrt{3}/2} = \mathbf{-\frac{2}{\sqrt{3}}} \]

4. Find \(\tan x\):

\[ \tan x = \frac{\sin x}{\cos x} = \frac{-\sqrt{3}/2}{-1/2} = \mathbf{\sqrt{3}} \]

5. Find \(\cot x\):

\[ \cot x = \frac{1}{\tan x} = \mathbf{\frac{1}{\sqrt{3}}} \]

Answer:

\[ \sin x = -\frac{\sqrt{3}}{2}, \text{cosec } x = -\frac{2}{\sqrt{3}}, \sec x = -2, \tan x = \sqrt{3}, \cot x = \frac{1}{\sqrt{3}} \]

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Question 2: Find the values of other five trigonometric functions if \(\sin x = \frac{3}{5}\), \(x\) lies in the second quadrant.

Solution:

Given: \(\sin x = \frac{3}{5}\).

Since \(x\) is in the 2nd quadrant, \(\sin x\) and \(\text{cosec } x\) are positive; others are negative.

1. Find \(\text{cosec } x\):

\[ \text{cosec } x = \frac{1}{\sin x} = \frac{1}{3/5} = \mathbf{\frac{5}{3}} \]

2. Find \(\cos x\) using \(\sin^2 x + \cos^2 x = 1\):

\[ \cos^2 x = 1 - \sin^2 x = 1 - \left(\frac{3}{5}\right)^2 = 1 - \frac{9}{25} = \frac{16}{25} \]

\[ \cos x = \pm \frac{4}{5} \]

Since \(x\) is in the 2nd quadrant, \(\cos x\) is negative:

\[ \cos x = \mathbf{-\frac{4}{5}} \]

3. Find \(\sec x\):

\[ \sec x = \frac{1}{\cos x} = \frac{1}{-4/5} = \mathbf{-\frac{5}{4}} \]

4. Find \(\tan x\):

\[ \tan x = \frac{\sin x}{\cos x} = \frac{3/5}{-4/5} = \mathbf{-\frac{3}{4}} \]

5. Find \(\cot x\):

\[ \cot x = \frac{1}{\tan x} = \frac{1}{-3/4} = \mathbf{-\frac{4}{3}} \]

Answer:

\[ \cos x = -\frac{4}{5}, \sec x = -\frac{5}{4}, \text{cosec } x = \frac{5}{3}, \tan x = -\frac{3}{4}, \cot x = -\frac{4}{3} \]

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Question 3: Find the values of other five trigonometric functions if \(\cot x = \frac{3}{4}\), \(x\) lies in the third quadrant.

Solution:

Given: \(\cot x = \frac{3}{4}\).

Since \(x\) is in the 3rd quadrant, \(\tan x\) and \(\cot x\) are positive; others are negative.

1. Find \(\tan x\):

\[ \tan x = \frac{1}{\cot x} = \frac{1}{3/4} = \mathbf{\frac{4}{3}} \]

2. Find \(\sec x\) using \(\sec^2 x = 1 + \tan^2 x\):

\[ \sec^2 x = 1 + \left(\frac{4}{3}\right)^2 = 1 + \frac{16}{9} = \frac{25}{9} \]

\[ \sec x = \pm \frac{5}{3} \]

Since \(x\) is in the 3rd quadrant, \(\sec x\) is negative:

\[ \sec x = \mathbf{-\frac{5}{3}} \]

3. Find \(\cos x\):

\[ \cos x = \frac{1}{\sec x} = \frac{1}{-5/3} = \mathbf{-\frac{3}{5}} \]

4. Find \(\sin x\) using \(\tan x = \frac{\sin x}{\cos x}\):

\[ \sin x = \tan x \cdot \cos x = \frac{4}{3} \cdot \left(-\frac{3}{5}\right) = \mathbf{-\frac{4}{5}} \]

5. Find \(\text{cosec } x\):

\[ \text{cosec } x = \frac{1}{\sin x} = \frac{1}{-4/5} = \mathbf{-\frac{5}{4}} \]

Answer:

\[ \sin x = -\frac{4}{5}, \cos x = -\frac{3}{5}, \tan x = \frac{4}{3}, \sec x = -\frac{5}{3}, \text{cosec } x = -\frac{5}{4} \]

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Question 4: Find the values of other five trigonometric functions if \(\sec x = \frac{13}{5}\), \(x\) lies in the fourth quadrant.

Solution:

Given: \(\sec x = \frac{13}{5}\).

Since \(x\) is in the 4th quadrant, \(\cos x\) and \(\sec x\) are positive; others are negative.

1. Find \(\cos x\):

\[ \cos x = \frac{1}{\sec x} = \mathbf{\frac{5}{13}} \]

2. Find \(\tan x\) using \(\sec^2 x = 1 + \tan^2 x\):

\[ \tan^2 x = \sec^2 x - 1 = \left(\frac{13}{5}\right)^2 - 1 = \frac{169}{25} - 1 = \frac{144}{25} \]

\[ \tan x = \pm \frac{12}{5} \]

Since \(x\) is in the 4th quadrant, \(\tan x\) is negative:

\[ \tan x = \mathbf{-\frac{12}{5}} \]

3. Find \(\cot x\):

\[ \cot x = \frac{1}{\tan x} = \mathbf{-\frac{5}{12}} \]

4. Find \(\sin x\) using \(\sin x = \tan x \cdot \cos x\):

\[ \sin x = \left(-\frac{12}{5}\right) \cdot \frac{5}{13} = \mathbf{-\frac{12}{13}} \]

5. Find \(\text{cosec } x\):

\[ \text{cosec } x = \frac{1}{\sin x} = \mathbf{-\frac{13}{12}} \]

Answer:

\[ \sin x = -\frac{12}{13}, \cos x = \frac{5}{13}, \tan x = -\frac{12}{5}, \cot x = -\frac{5}{12}, \text{cosec } x = -\frac{13}{12} \]

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Question 5: Find the values of other five trigonometric functions if \(\tan x = -\frac{5}{12}\), \(x\) lies in the second quadrant.

Solution:

Given: \(\tan x = -\frac{5}{12}\).

Since \(x\) is in the 2nd quadrant, \(\sin x\) and \(\text{cosec } x\) are positive; others are negative.

1. Find \(\cot x\):

\[ \cot x = \frac{1}{\tan x} = \mathbf{-\frac{12}{5}} \]

2. Find \(\sec x\) using \(\sec^2 x = 1 + \tan^2 x\):

\[ \sec^2 x = 1 + \left(-\frac{5}{12}\right)^2 = 1 + \frac{25}{144} = \frac{169}{144} \]

\[ \sec x = \pm \frac{13}{12} \]

Since \(x\) is in the 2nd quadrant, \(\sec x\) is negative:

\[ \sec x = \mathbf{-\frac{13}{12}} \]

3. Find \(\cos x\):

\[ \cos x = \frac{1}{\sec x} = \mathbf{-\frac{12}{13}} \]

4. Find \(\sin x\) using \(\sin x = \tan x \cdot \cos x\):

\[ \sin x = \left(-\frac{5}{12}\right) \cdot \left(-\frac{12}{13}\right) = \mathbf{\frac{5}{13}} \]

5. Find \(\text{cosec } x\):

\[ \text{cosec } x = \frac{1}{\sin x} = \mathbf{\frac{13}{5}} \]

Answer:

\[ \sin x = \frac{5}{13}, \cos x = -\frac{12}{13}, \cot x = -\frac{12}{5}, \sec x = -\frac{13}{12}, \text{cosec } x = \frac{13}{5} \]

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Question 6: Find the value of \(\sin 765^{\circ}\).

Solution:

We know that the sine function repeats after \(360^{\circ}\).

We can write \(765^{\circ}\) as multiples of \(360^{\circ}\):

\[ 765^{\circ} = 2 \times 360^{\circ} + 45^{\circ} \]

\[ \therefore \sin 765^{\circ} = \sin (2 \times 360^{\circ} + 45^{\circ}) = \sin 45^{\circ} \]

\[ = \mathbf{\frac{1}{\sqrt{2}}} \]

Answer:

\[ \frac{1}{\sqrt{2}} \]

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Question 7: Find the value of \(\text{cosec } (-1410^{\circ})\).

Solution:

We know that \(\text{cosec } (-x) = -\text{cosec } x\).

\[ \text{cosec } (-1410^{\circ}) = -\text{cosec } (1410^{\circ}) \]

We can write \(1410^{\circ}\) as:

\[ 1410^{\circ} = 4 \times 360^{\circ} - 30^{\circ} \]

Alternatively, we can add multiples of \(360^{\circ}\) to the angle until it becomes positive.

\[ -1410^{\circ} + 4 \times 360^{\circ} = -1410^{\circ} + 1440^{\circ} = 30^{\circ} \]

So,

\[ \text{cosec } (-1410^{\circ}) = \text{cosec } 30^{\circ} = \mathbf{2} \]

Answer:

2

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Question 8: Find the value of \(\tan \frac{19\pi}{3}\).

Solution:

We know that \(\tan\) repeats after \(\pi\). Let's convert the fraction:

\[ \frac{19\pi}{3} = 6\pi + \frac{\pi}{3} \]

\[ \tan \frac{19\pi}{3} = \tan \left(6\pi + \frac{\pi}{3}\right) \]

Since \(6\pi\) is a multiple of \(\pi\), the value is simply:

\[ = \tan \frac{\pi}{3} = \tan 60^{\circ} = \mathbf{\sqrt{3}} \]

Answer:

\[ \sqrt{3} \]

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Question 9: Find the value of \(\sin \left(-\frac{11\pi}{3}\right)\).

Solution:

We know that \(\sin(-x) = -\sin x\).

\[ \sin \left(-\frac{11\pi}{3}\right) = -\sin \frac{11\pi}{3} \]

Let's simplify \(\frac{11\pi}{3}\):

\[ \frac{11\pi}{3} = 4\pi - \frac{\pi}{3} \]

\[ \sin \left(4\pi - \frac{\pi}{3}\right) = -\sin \frac{\pi}{3} \]

(Since \(4\pi - \theta\) is in the 4th quadrant where sin is negative)

Substituting back:

\[ -\sin \frac{11\pi}{3} = - \left(-\sin \frac{\pi}{3}\right) = \sin \frac{\pi}{3} \]

\[ = \mathbf{\frac{\sqrt{3}}{2}} \]

Answer:

\[ \frac{\sqrt{3}}{2} \]

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Question 10: Find the value of \(\cot \left(-\frac{15\pi}{4}\right)\).

Solution:

We know \(\cot(-x) = -\cot x\).

\[ \cot \left(-\frac{15\pi}{4}\right) = -\cot \frac{15\pi}{4} \]

Simplify \(\frac{15\pi}{4}\):

\[ \frac{15\pi}{4} = 4\pi - \frac{\pi}{4} \]

\[ \cot \left(4\pi - \frac{\pi}{4}\right) = -\cot \frac{\pi}{4} \]

(Since \(4\pi - \theta\) is in the 4th quadrant where cot is negative)

Substituting back:

\[ -\cot \frac{15\pi}{4} = - \left(-\cot \frac{\pi}{4}\right) = \cot \frac{\pi}{4} \]

\[ = \cot 45^{\circ} = \mathbf{1} \]

Answer:

1

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Expert Tips & Common Mistakes

- Quadrant Signs: Always double check the quadrant. A common mistake is calculating the value correctly (e.g., 4/5) but forgetting the minus sign.

- Negative Angles: Remember that \(\cos(-x) = \cos x\) and \(\sec(-x) = \sec x\) (they "absorb" the negative sign). All other functions pull the negative sign out (e.g., \(\sin(-x) = -\sin x\)).

- Full Rotations: For large angles like \(765^{\circ}\) or \(-1410^{\circ}\), simply add or subtract \(360^{\circ}\) (or \(2\pi\)) until the angle is small and easy to work with.