NCERT Class 10 Mathematics — Introduction to Trigonometry, Exercise 8.2. All 4 questions solved.

Exercise 8.2 is where the standard angles arrive. Every question is settled by one table — the ratios of $0^\circ$, $30^\circ$, $45^\circ$, $60^\circ$ and $90^\circ$, which the chapter derives from an equilateral triangle cut in half and from an isosceles right triangle:

Once the right values are in place, the rest is fraction arithmetic and, in question 1(iii) and (iv), rationalising a denominator.

Key insight. Only one row needs remembering. The sines of $0^\circ$, $30^\circ$, $45^\circ$, $60^\circ$, $90^\circ$ are $\frac{\sqrt0}{2}, \frac{\sqrt1}{2}, \frac{\sqrt2}{2}, \frac{\sqrt3}{2}, \frac{\sqrt4}{2}$; the cosines are the same row read backwards; $\tan A = \frac{\sin A}{\cos A}$; and the last three rows are reciprocals. Nearly every wrong answer in this exercise is a wrong table value — $\operatorname{cosec} 60^\circ$ confused with $\sec 60^\circ$, or $\tan 30^\circ$ with $\tan 60^\circ$ — not a wrong method.

Question 1

Evaluate the following:

(i) $\sin 60^\circ \cos 30^\circ + \sin 30^\circ \cos 60^\circ$

(ii) $2\tan^2 45^\circ + \cos^2 30^\circ – \sin^2 60^\circ$

(iii) $\dfrac{\cos 45^\circ}{\sec 30^\circ + \operatorname{cosec} 30^\circ}$

(iv) $\dfrac{\sin 30^\circ + \tan 45^\circ – \operatorname{cosec} 60^\circ}{\sec 30^\circ + \cos 60^\circ + \cot 45^\circ}$

(v) $\dfrac{5\cos^2 60^\circ + 4\sec^2 30^\circ – \tan^2 45^\circ}{\sin^2 30^\circ + \cos^2 30^\circ}$

Solution. Substitute the table values, then simplify. A squared ratio such as $\tan^2 45^\circ$ means $(\tan 45^\circ)^2$ — square the value, not the angle.

(i)

$$\frac{\sqrt3}{2} \cdot \frac{\sqrt3}{2} + \frac12 \cdot \frac12 = \frac34 + \frac14 = 1$$

(ii) $\cos 30^\circ$ and $\sin 60^\circ$ are both $\frac{\sqrt3}{2}$, so the last two terms cancel before any arithmetic:

$$2(1)^2 + \left(\frac{\sqrt3}{2}\right)^2 – \left(\frac{\sqrt3}{2}\right)^2 = 2 + \frac34 – \frac34 = 2$$

(iii) Combine the denominator into a single fraction first, because only a single fraction can be inverted:

$$\sec 30^\circ + \operatorname{cosec} 30^\circ = \frac{2}{\sqrt3} + 2 = \frac{2 + 2\sqrt3}{\sqrt3} = \frac{2(\sqrt3 + 1)}{\sqrt3}$$

$$\frac{\cos 45^\circ}{\sec 30^\circ + \operatorname{cosec} 30^\circ} = \frac{1}{\sqrt2} \cdot \frac{\sqrt3}{2(\sqrt3 + 1)} = \frac{\sqrt3}{2\sqrt2(\sqrt3 + 1)}$$

Rationalise by multiplying top and bottom by $\sqrt3 – 1$, which turns $(\sqrt3 + 1)(\sqrt3 – 1)$ into $3 – 1 = 2$:

$$\frac{\sqrt3(\sqrt3 – 1)}{2\sqrt2 \cdot 2} = \frac{3 – \sqrt3}{4\sqrt2} = \frac{(3 – \sqrt3)\sqrt2}{8} = \frac{3\sqrt2 – \sqrt6}{8}$$

(iv) Put numerator and denominator over the common denominator $2\sqrt3$:

$$\text{Numerator} = \frac12 + 1 – \frac{2}{\sqrt3} = \frac32 – \frac{2}{\sqrt3} = \frac{3\sqrt3 – 4}{2\sqrt3}, \qquad \text{Denominator} = \frac{2}{\sqrt3} + \frac12 + 1 = \frac{3\sqrt3 + 4}{2\sqrt3}$$

The $2\sqrt3$ cancels, leaving $\dfrac{3\sqrt3 – 4}{3\sqrt3 + 4}$. Multiply top and bottom by $3\sqrt3 – 4$:

$$\frac{(3\sqrt3 – 4)^2}{(3\sqrt3)^2 – 4^2} = \frac{27 – 24\sqrt3 + 16}{27 – 16} = \frac{43 – 24\sqrt3}{11}$$

(v) The denominator is $\frac14 + \frac34 = 1$, so only the numerator needs work:

$$5\left(\frac12\right)^2 + 4\left(\frac{2}{\sqrt3}\right)^2 – 1^2 = \frac54 + \frac{16}{3} – 1 = \frac{15 + 64 – 12}{12} = \frac{67}{12}$$

Question 2

Choose the correct option and justify your choice:

(i) $\dfrac{2\tan 30^\circ}{1 + \tan^2 30^\circ} =$

With $\tan 30^\circ = \frac{1}{\sqrt3}$ and $\tan^2 30^\circ = \frac13$:

$$\frac{\frac{2}{\sqrt3}}{1 + \frac13} = \frac{2}{\sqrt3} \cdot \frac34 = \frac{3}{2\sqrt3} = \frac{\sqrt3}{2} = \sin 60^\circ$$

The other options are $\frac12$, $\sqrt3$ and $\frac12$, none of which is $\frac{\sqrt3}{2}$. (A).

(ii) $\dfrac{1 – \tan^2 45^\circ}{1 + \tan^2 45^\circ} =$

$\tan 45^\circ = 1$, so the numerator is $1 – 1 = 0$ while the denominator is $2$:

$$\frac{1 – 1}{1 + 1} = \frac{0}{2} = 0$$

(D). Option (A) is not even a number, since $\tan 90^\circ$ is not defined.

(iii) $\sin 2A = 2\sin A$ is true when $A =$

This is not an identity, so test the options. At $A = 0^\circ$: $\sin 0^\circ = 0$ and $2\sin 0^\circ = 0$, so it holds. The others all fail:

• $A = 30^\circ$: $\sin 60^\circ = \frac{\sqrt3}{2}$, but $2\sin 30^\circ = 1$.
• $A = 45^\circ$: $2\sin 45^\circ = \sqrt2 > 1$, and no sine exceeds $1$.
• $A = 60^\circ$: $2\sin 60^\circ = \sqrt3 > 1$, which rules it out the same way.

(A).

(iv) $\dfrac{2\tan 30^\circ}{1 – \tan^2 30^\circ} =$

The same numerator as part (i), but now the denominator is $1 – \frac13 = \frac23$:

$$\frac{\frac{2}{\sqrt3}}{\frac23} = \frac{2}{\sqrt3} \cdot \frac32 = \frac{3}{\sqrt3} = \sqrt3 = \tan 60^\circ$$

(C).

Parts (i), (ii) and (iv) are Class 11’s double-angle formulae in disguise — $\sin 2A$, $\cos 2A$ and $\tan 2A$ written in terms of $\tan A$ — which is why $A = 30^\circ$ produces ratios of $60^\circ$, and $A = 45^\circ$ produces $\cos 90^\circ = 0$.

Question 3

If $\tan(A + B) = \sqrt3$ and $\tan(A – B) = \dfrac{1}{\sqrt3}$; $0^\circ < A + B \le 90^\circ$; $A > B$, find $A$ and $B$.

Solution. The conditions are there so that each tangent value names exactly one angle. Between $0^\circ$ and $90^\circ$ the tangent keeps increasing, so it takes each value only once.

Since $0^\circ < A + B \le 90^\circ$ and $\tan 60^\circ = \sqrt3$,

$$A + B = 60^\circ$$

Since $A > B$, the angle $A – B$ is positive and smaller than $A + B$, so it too lies between $0^\circ$ and $90^\circ$. As $\tan 30^\circ = \frac{1}{\sqrt3}$,

$$A – B = 30^\circ$$

Adding the two equations gives $2A = 90^\circ$, so $A = 45^\circ$; subtracting them gives $2B = 30^\circ$, so $B = 15^\circ$.

Check: $\tan(45^\circ + 15^\circ) = \tan 60^\circ = \sqrt3$ and $\tan(45^\circ – 15^\circ) = \tan 30^\circ = \frac{1}{\sqrt3}$.

Question 4

State whether the following are true or false. Justify your answer.

(i) $\sin(A + B) = \sin A + \sin B$. (ii) The value of $\sin\theta$ increases as $\theta$ increases. (iii) The value of $\cos\theta$ increases as $\theta$ increases. (iv) $\sin\theta = \cos\theta$ for all values of $\theta$. (v) $\cot A$ is not defined for $A = 0^\circ$.

Solution. One counterexample is enough to make a “for all” statement false; a true statement needs a reason that covers every case.

(i) False. Take $A = 60^\circ$ and $B = 30^\circ$. Then $\sin(A + B) = \sin 90^\circ = 1$, but $\sin 60^\circ + \sin 30^\circ = \frac{\sqrt3}{2} + \frac12 = \frac{\sqrt3 + 1}{2} \approx 1.366$. “$\sin$” is the name of a ratio, not a number multiplying the bracket, so it does not distribute over a sum.

(ii) True. Down the table, as $\theta$ goes $0^\circ, 30^\circ, 45^\circ, 60^\circ, 90^\circ$, $\sin\theta$ goes $0, 0.5, 0.707, 0.866, 1$. The geometric reason: keep the hypotenuse fixed and open the angle, and the opposite side grows. (This is within $0^\circ$ to $90^\circ$, the range this chapter works in; beyond $90^\circ$ sine begins to fall, which Class 11 takes up.)

(iii) False. Over the same angles $\cos\theta$ goes $1, 0.866, 0.707, 0.5, 0$ — it decreases. With the hypotenuse fixed, the adjacent side shrinks as the angle opens.

(iv) False. At $\theta = 0^\circ$, $\sin 0^\circ = 0$ but $\cos 0^\circ = 1$. The two are equal only at $\theta = 45^\circ$, where the right triangle is isosceles.

(v) True. $\cot A = \dfrac{\cos A}{\sin A}$, and $\sin 0^\circ = 0$, so $\cot 0^\circ$ would be $\dfrac10$, which is not defined.

Common mistakes

• Question 1(ii) and (v), squaring the angle. $\tan^2 45^\circ$ is $(\tan 45^\circ)^2 = 1$, and $\sec^2 30^\circ$ is $\left(\frac{2}{\sqrt3}\right)^2 = \frac43$. The notation puts the $2$ next to the ratio precisely so that it is not read as belonging to the angle.
• Question 1(iii), inverting term by term. Writing $\frac{1}{\sec 30^\circ + \operatorname{cosec} 30^\circ}$ as $\cos 30^\circ + \sin 30^\circ$ treats the reciprocal of a sum as the sum of reciprocals, which it is not. Combine the denominator into one fraction first.
• Question 1(iv), taking $\operatorname{cosec} 60^\circ = 2$. That is $\sec 60^\circ$. Cosecant pairs with sine, so $\operatorname{cosec} 60^\circ = \frac{1}{\sin 60^\circ} = \frac{2}{\sqrt3}$.
• Question 2(i) against (iv). The two expressions differ only in the sign in the denominator, $1 + \tan^2 30^\circ$ against $1 – \tan^2 30^\circ$, and the answers differ accordingly ($\sin 60^\circ$ against $\tan 60^\circ$). Copying the wrong sign swaps the options.
• Question 2(iii), “taking the 2 out”. $\sin 2A = 2\sin A$ is not an identity; it holds at $0^\circ$ and at no other option. This is the same misconception as question 4(i) — treating $\sin$ as a multiplier.
• Question 3, swapping $\tan 30^\circ$ and $\tan 60^\circ$. $\sqrt3$ is the larger value, so it belongs to the larger angle: $\tan 60^\circ = \sqrt3$. Getting this backwards gives $A + B = 30^\circ$ and $A – B = 60^\circ$, which would make $B$ negative — a sign that something has gone wrong.
• Question 4(ii), answering false because sine falls after $90^\circ$. In this chapter $\theta$ runs from $0^\circ$ to $90^\circ$, and there sine does increase.

Practise next

• Exercise 8.3 — the identities, where a value from the table above is the quickest way to test any identity before proving it.
• Exercise 8.1 — revise the ratios from a triangle; its question 9 is the $30^\circ$–$60^\circ$–$90^\circ$ triangle in disguise.
• Chapter 9, Exercise 9.1 — heights and distances, which use almost nothing but $\tan 30^\circ$, $\tan 45^\circ$ and $\tan 60^\circ$ from this table.