NCERT Class 11 Mathematics — Trigonometric Functions, Exercise 3.2. All 10 questions solved.

Two tools do all the work here. The first is the Pythagorean identity and its companions:

$$\sin^2 x + \cos^2 x = 1, \qquad 1 + \tan^2 x = \sec^2 x, \qquad 1 + \cot^2 x = \operatorname{cosec}^2 x$$

The second is the sign chart — which functions are positive in which quadrant. The usual mnemonic is A-S-T-C (All, Sine, Tangent, Cosine positive in quadrants I, II, III, IV respectively), read anticlockwise from the first quadrant.

The reciprocal functions carry the sign of the function they invert.

For questions 6 to 10, periodicity does the reducing: $\sin$, $\cos$, $\operatorname{cosec}$ and $\sec$ repeat every $2\pi$ (or $360^\circ$); $\tan$ and $\cot$ repeat every $\pi$.

Key insight. The identity fixes the magnitude; the quadrant fixes the sign. Doing these in the wrong order is where the errors come from — solve $\sin^2 x = \tfrac34$ to get $|\sin x| = \tfrac{\sqrt3}{2}$, and only then let the quadrant decide whether to write $+$ or $-$. A useful check on every answer: $\sin x$ and $\operatorname{cosec} x$ must share a sign, as must $\cos x$ with $\sec x$, and $\tan x$ with $\cot x$.

Find the values of the other five trigonometric functions in questions 1 to 5.

Question 1

$\cos x = -\dfrac{1}{2}$, $x$ lies in the third quadrant.

Solution. From $\sin^2 x = 1 – \cos^2 x$:

$$\sin^2 x = 1 – \frac{1}{4} = \frac{3}{4} \quad\Longrightarrow\quad |\sin x| = \frac{\sqrt3}{2}$$

In the third quadrant $\sin x$ is negative, so $\sin x = -\dfrac{\sqrt3}{2}$.

The rest follow by definition:

$$\operatorname{cosec} x = \frac{1}{\sin x} = -\frac{2}{\sqrt3}, \qquad \sec x = \frac{1}{\cos x} = -2$$

$$\tan x = \frac{\sin x}{\cos x} = \frac{-\sqrt3/2}{-1/2} = \sqrt3, \qquad \cot x = \frac{1}{\sqrt3}$$

$\tan x$ comes out positive, as the third quadrant requires — a free check.

Question 2

$\sin x = \dfrac{3}{5}$, $x$ lies in the second quadrant.

Solution.

$$\cos^2 x = 1 – \frac{9}{25} = \frac{16}{25} \quad\Longrightarrow\quad |\cos x| = \frac{4}{5}$$

In the second quadrant $\cos x$ is negative, so $\cos x = -\dfrac{4}{5}$.

$$\operatorname{cosec} x = \frac{5}{3}, \qquad \sec x = -\frac{5}{4}$$

$$\tan x = \frac{3/5}{-4/5} = -\frac{3}{4}, \qquad \cot x = -\frac{4}{3}$$

Question 3

$\cot x = \dfrac{3}{4}$, $x$ lies in the third quadrant.

Solution. Use $\operatorname{cosec}^2 x = 1 + \cot^2 x$:

$$\operatorname{cosec}^2 x = 1 + \frac{9}{16} = \frac{25}{16} \quad\Longrightarrow\quad |\operatorname{cosec} x| = \frac{5}{4}$$

In the third quadrant $\sin x$, and hence $\operatorname{cosec} x$, is negative:

$$\operatorname{cosec} x = -\frac{5}{4}, \qquad \sin x = -\frac{4}{5}$$

Then $\tan x = \dfrac{1}{\cot x} = \dfrac{4}{3}$, and

$$\cos x = \sin x \cdot \cot x = \left(-\frac45\right)\left(\frac34\right) = -\frac{3}{5}, \qquad \sec x = -\frac{5}{3}$$

Question 4

$\sec x = \dfrac{13}{5}$, $x$ lies in the fourth quadrant.

Solution. $\cos x = \dfrac{5}{13}$, which is positive — consistent with the fourth quadrant, where cosine is the positive one.

$$\sin^2 x = 1 – \frac{25}{169} = \frac{144}{169} \quad\Longrightarrow\quad |\sin x| = \frac{12}{13}$$

In the fourth quadrant $\sin x$ is negative:

$$\sin x = -\frac{12}{13}, \qquad \operatorname{cosec} x = -\frac{13}{12}$$

$$\tan x = \frac{-12/13}{5/13} = -\frac{12}{5}, \qquad \cot x = -\frac{5}{12}$$

Question 5

$\tan x = -\dfrac{5}{12}$, $x$ lies in the second quadrant.

Solution. From $\sec^2 x = 1 + \tan^2 x$:

$$\sec^2 x = 1 + \frac{25}{144} = \frac{169}{144} \quad\Longrightarrow\quad |\sec x| = \frac{13}{12}$$

In the second quadrant $\cos x$ and $\sec x$ are negative:

$$\sec x = -\frac{13}{12}, \qquad \cos x = -\frac{12}{13}$$

Then

$$\sin x = \tan x \cdot \cos x = \left(-\frac{5}{12}\right)\left(-\frac{12}{13}\right) = \frac{5}{13}$$

which is positive, as the second quadrant requires. Finally

$$\operatorname{cosec} x = \frac{13}{5}, \qquad \cot x = -\frac{12}{5}$$

Find the values of the trigonometric functions in questions 6 to 10.

Question 6

$\sin 765^\circ$

Solution. Sine has period $360^\circ$, so subtract whole turns until the angle lands in $[0^\circ, 360^\circ)$:

$$765^\circ – 2 \times 360^\circ = 765^\circ – 720^\circ = 45^\circ$$

$$\sin 765^\circ = \sin 45^\circ = \frac{1}{\sqrt2}$$

Question 7

$\operatorname{cosec}(-1410^\circ)$

Solution. Add whole turns until the angle is in range. Four turns is $1440^\circ$:

$$-1410^\circ + 1440^\circ = 30^\circ$$

$$\operatorname{cosec}(-1410^\circ) = \operatorname{cosec} 30^\circ = \frac{1}{\sin 30^\circ} = \frac{1}{1/2} = 2$$

Adding a multiple of $360^\circ$ to a negative angle is legitimate for the same reason subtracting one is — a full turn returns to the same point on the unit circle.

Question 8

$\tan \dfrac{19\pi}{3}$

Solution. Tangent has period $\pi$, but it is just as easy to use $2\pi$ here. Since $\dfrac{19\pi}{3} = 6\pi + \dfrac{\pi}{3}$, subtracting three full turns:

$$\tan \frac{19\pi}{3} = \tan \frac{\pi}{3} = \sqrt3$$

Question 9

$\sin\left(-\dfrac{11\pi}{3}\right)$

Solution. Add two full turns, $4\pi = \dfrac{12\pi}{3}$:

$$-\frac{11\pi}{3} + \frac{12\pi}{3} = \frac{\pi}{3}$$

$$\sin\left(-\frac{11\pi}{3}\right) = \sin\frac{\pi}{3} = \frac{\sqrt3}{2}$$

Question 10

$\cot\left(-\dfrac{15\pi}{4}\right)$

Solution. Add two full turns, $4\pi = \dfrac{16\pi}{4}$:

$$-\frac{15\pi}{4} + \frac{16\pi}{4} = \frac{\pi}{4}$$

$$\cot\left(-\frac{15\pi}{4}\right) = \cot\frac{\pi}{4} = 1$$

Common mistakes

• Taking the positive square root automatically. $\sin^2 x = \tfrac{3}{4}$ gives $\sin x = \pm\tfrac{\sqrt3}{2}$; only the quadrant decides. In question 1 the answer is negative, and a student who writes $+\tfrac{\sqrt3}{2}$ gets every subsequent sign wrong too.
• Question 3, inverting $\cot x$ into $\tan x$ and then forgetting the sign. $\cot x = \tfrac34$ gives $\tan x = \tfrac43$, both positive, which is right for the third quadrant — but $\sin x$ and $\cos x$ are both negative there, and it is easy to lose one of those minus signs when reconstructing them.
• Using $\sin^2 + \cos^2 = 1$ when $\tan$ or $\cot$ is given. It works, but $1 + \tan^2 = \sec^2$ and $1 + \cot^2 = \operatorname{cosec}^2$ get there in one step. Questions 3 and 5 are built for those two identities.
• Question 5, misreading which functions are negative in quadrant II. Sine is the survivor there: $\sin$ and $\operatorname{cosec}$ positive, everything else negative.
• Questions 7, 9 and 10, subtracting when the angle is negative. For a negative angle you must add multiples of $360^\circ$ (or $2\pi$) to bring it into range.
• Question 8, reducing modulo $\pi$ carelessly. $\dfrac{19\pi}{3}$ minus $6\pi$ is $\dfrac{\pi}{3}$, not $\dfrac{\pi}{6}$ — write $6\pi$ as $\dfrac{18\pi}{3}$ before subtracting.

Practise next

• Exercise 3.3 — the sum and difference identities, and the proofs built from them. The quadrant reasoning here reappears whenever a half-angle sign has to be decided.
• Exercise 3.1 — worth revising if converting $\dfrac{19\pi}{3}$ or $-1410^\circ$ into a standard angle felt slow.