NCERT Class 10 Mathematics — Introduction to Trigonometry, Exercise 8.3. All 4 questions solved.

Exercise 8.3 is the identities exercise. Three identities, all from Pythagoras’ theorem, carry every question:

$$\sin^2 A + \cos^2 A = 1$$

$$1 + \tan^2 A = \sec^2 A \quad (0^\circ \le A < 90^\circ), \qquad 1 + \cot^2 A = \operatorname{cosec}^2 A \quad (0^\circ < A \le 90^\circ)$$

The second and third are the first divided through by $\cos^2 A$ and by $\sin^2 A$. Beyond them, two habits do most of the work: when stuck, write everything in $\sin$ and $\cos$; and when a fraction has $1 \pm \sin A$ or $1 \pm \cos A$ in it, multiply by the other sign to make a difference of squares.

Key insight. Each identity is a difference of two squares that equals $1$: $$\operatorname{cosec}^2 A – \cot^2 A = (\operatorname{cosec} A – \cot A)(\operatorname{cosec} A + \cot A) = 1$$ and likewise $\sec^2 A – \tan^2 A = 1$ and $(1 – \sin A)(1 + \sin A) = \cos^2 A$. Replacing a plain $1$ by such a product — or a product by $1$ — is the move that unlocks the harder proofs: question 4(v) turns on it, and so do 3(iii) and 4(vi).

Question 1

Express the trigonometric ratios $\sin A$, $\sec A$ and $\tan A$ in terms of $\cot A$.

Solution. $\tan A$ is immediate, since it is the reciprocal of $\cot A$:

$$\tan A = \frac{1}{\cot A}$$

For $\sin A$, use the identity that links $\cot A$ to a reciprocal of $\sin A$, namely $\operatorname{cosec}^2 A = 1 + \cot^2 A$. Taking the positive square root, because every ratio of an acute angle is positive,

$$\operatorname{cosec} A = \sqrt{1 + \cot^2 A} \quad\Longrightarrow\quad \sin A = \frac{1}{\sqrt{1 + \cot^2 A}}$$

For $\sec A$, write $\cos A = \cot A \cdot \sin A$ (since $\cot A = \frac{\cos A}{\sin A}$):

$$\cos A = \frac{\cot A}{\sqrt{1 + \cot^2 A}} \quad\Longrightarrow\quad \sec A = \frac{1}{\cos A} = \frac{\sqrt{1 + \cot^2 A}}{\cot A}$$

Question 2

Write all the other trigonometric ratios of $\angle A$ in terms of $\sec A$.

Solution. $\cos A$ is simply the reciprocal:

$$\cos A = \frac{1}{\sec A}$$

The identity containing $\sec A$ is $1 + \tan^2 A = \sec^2 A$, which gives $\tan A$ directly (positive root, as $A$ is acute):

$$\tan A = \sqrt{\sec^2 A – 1}$$

Every other ratio now follows from these two. $\sin A = \tan A \cdot \cos A$, and $\cot A$ and $\operatorname{cosec} A$ are reciprocals of $\tan A$ and $\sin A$:

$$\sin A = \frac{\sqrt{\sec^2 A – 1}}{\sec A}, \qquad \cot A = \frac{1}{\sqrt{\sec^2 A – 1}}, \qquad \operatorname{cosec} A = \frac{\sec A}{\sqrt{\sec^2 A – 1}}$$

Question 3

Choose the correct option. Justify your choice.

(i) $9\sec^2 A – 9\tan^2 A =$

Take out the common factor and use $\sec^2 A – \tan^2 A = 1$:

$$9(\sec^2 A – \tan^2 A) = 9 \times 1 = 9$$

(B).

(ii) $(1 + \tan\theta + \sec\theta)(1 + \cot\theta – \operatorname{cosec}\theta) =$

Write each bracket over a single denominator using $\sin$ and $\cos$:

$$\left(\frac{\cos\theta + \sin\theta + 1}{\cos\theta}\right)\left(\frac{\sin\theta + \cos\theta – 1}{\sin\theta}\right)$$

The numerators are $(\sin\theta + \cos\theta) + 1$ and $(\sin\theta + \cos\theta) – 1$, a sum and difference, so their product is a difference of squares:

$$\frac{(\sin\theta + \cos\theta)^2 – 1}{\sin\theta\cos\theta} = \frac{\sin^2\theta + \cos^2\theta + 2\sin\theta\cos\theta – 1}{\sin\theta\cos\theta} = \frac{2\sin\theta\cos\theta}{\sin\theta\cos\theta} = 2$$

(C).

(iii) $(\sec A + \tan A)(1 – \sin A) =$

$\sec A + \tan A = \dfrac{1 + \sin A}{\cos A}$, and $1 + \sin A$ meets $1 – \sin A$ to make $1 – \sin^2 A = \cos^2 A$:

$$\frac{(1 + \sin A)(1 – \sin A)}{\cos A} = \frac{\cos^2 A}{\cos A} = \cos A$$

(D).

(iv) $\dfrac{1 + \tan^2 A}{1 + \cot^2 A} =$

Numerator and denominator are each one of the identities:

$$\frac{\sec^2 A}{\operatorname{cosec}^2 A} = \frac{1/\cos^2 A}{1/\sin^2 A} = \frac{\sin^2 A}{\cos^2 A} = \tan^2 A$$

(D). Option (B) could never be right: a ratio of two positive quantities cannot be negative.

Question 4

Prove the following identities, where the angles involved are acute angles for which the expressions are defined.

Question 4 (i)

$$(\operatorname{cosec}\theta – \cot\theta)^2 = \frac{1 – \cos\theta}{1 + \cos\theta}$$

Solution. The right-hand side has only $\cos\theta$, so bring the left-hand side to $\sin$ and $\cos$ and then turn $\sin^2\theta$ into cosines:

$$\text{LHS} = \left(\frac{1}{\sin\theta} – \frac{\cos\theta}{\sin\theta}\right)^2 = \frac{(1 – \cos\theta)^2}{\sin^2\theta}$$

Now $\sin^2\theta = 1 – \cos^2\theta = (1 – \cos\theta)(1 + \cos\theta)$, so one factor $1 – \cos\theta$ cancels (it is not zero, since $\theta$ is acute):

$$\frac{(1 – \cos\theta)^2}{(1 – \cos\theta)(1 + \cos\theta)} = \frac{1 – \cos\theta}{1 + \cos\theta} = \text{RHS}$$

Question 4 (ii)

$$\frac{\cos A}{1 + \sin A} + \frac{1 + \sin A}{\cos A} = 2\sec A$$

Solution. Add the two fractions over the common denominator $(1 + \sin A)\cos A$:

$$\text{LHS} = \frac{\cos^2 A + (1 + \sin A)^2}{(1 + \sin A)\cos A} = \frac{\cos^2 A + 1 + 2\sin A + \sin^2 A}{(1 + \sin A)\cos A}$$

The $\cos^2 A + \sin^2 A$ in the numerator is $1$, which leaves $2 + 2\sin A$ — a multiple of the factor $1 + \sin A$ already in the denominator:

$$\frac{2(1 + \sin A)}{(1 + \sin A)\cos A} = \frac{2}{\cos A} = 2\sec A = \text{RHS}$$

Question 4 (iii)

$$\frac{\tan\theta}{1 – \cot\theta} + \frac{\cot\theta}{1 – \tan\theta} = 1 + \sec\theta\operatorname{cosec}\theta$$

[Hint: Write the expression in terms of $\sin\theta$ and $\cos\theta$.]

Solution. Following the hint, simplify each term separately:

$$\frac{\tan\theta}{1 – \cot\theta} = \frac{\sin\theta/\cos\theta}{(\sin\theta – \cos\theta)/\sin\theta} = \frac{\sin^2\theta}{\cos\theta(\sin\theta – \cos\theta)}$$

$$\frac{\cot\theta}{1 – \tan\theta} = \frac{\cos\theta/\sin\theta}{(\cos\theta – \sin\theta)/\cos\theta} = \frac{\cos^2\theta}{\sin\theta(\cos\theta – \sin\theta)} = -\frac{\cos^2\theta}{\sin\theta(\sin\theta – \cos\theta)}$$

The second denominator contains $\cos\theta – \sin\theta$, the negative of the first one’s $\sin\theta – \cos\theta$; pulling out that minus sign is what lets the two fractions share a denominator. Over $\sin\theta\cos\theta(\sin\theta – \cos\theta)$:

$$\text{LHS} = \frac{\sin^3\theta – \cos^3\theta}{\sin\theta\cos\theta(\sin\theta – \cos\theta)}$$

Factorise the difference of cubes, $a^3 – b^3 = (a – b)(a^2 + ab + b^2)$, and cancel $\sin\theta – \cos\theta$ (non-zero, because the expression is undefined at $\theta = 45^\circ$):

$$\frac{(\sin\theta – \cos\theta)(\sin^2\theta + \sin\theta\cos\theta + \cos^2\theta)}{\sin\theta\cos\theta(\sin\theta – \cos\theta)} = \frac{1 + \sin\theta\cos\theta}{\sin\theta\cos\theta}$$

Split the fraction:

$$\frac{1}{\sin\theta\cos\theta} + 1 = 1 + \sec\theta\operatorname{cosec}\theta = \text{RHS}$$

Question 4 (iv)

$$\frac{1 + \sec A}{\sec A} = \frac{\sin^2 A}{1 – \cos A}$$

[Hint: Simplify LHS and RHS separately.]

Solution. Neither side turns easily into the other, so bring both to the same simple expression.

$$\text{LHS} = \frac{1}{\sec A} + \frac{\sec A}{\sec A} = \cos A + 1$$

$$\text{RHS} = \frac{1 – \cos^2 A}{1 – \cos A} = \frac{(1 – \cos A)(1 + \cos A)}{1 – \cos A} = 1 + \cos A$$

Both sides equal $1 + \cos A$, so they are equal to each other.

Question 4 (v)

$$\frac{\cos A – \sin A + 1}{\cos A + \sin A – 1} = \operatorname{cosec} A + \cot A, \text{ using the identity } \operatorname{cosec}^2 A = 1 + \cot^2 A.$$

Solution. The right-hand side is in $\operatorname{cosec}$ and $\cot$, and the suggested identity is in them too, so divide the numerator and denominator by $\sin A$ (non-zero for an acute angle) to bring them in:

$$\text{LHS} = \frac{\cot A – 1 + \operatorname{cosec} A}{\cot A + 1 – \operatorname{cosec} A}$$

Now use the identity in the form $1 = \operatorname{cosec}^2 A – \cot^2 A = (\operatorname{cosec} A – \cot A)(\operatorname{cosec} A + \cot A)$ to replace the $1$ in the numerator:

$$\text{Numerator} = (\operatorname{cosec} A + \cot A) – (\operatorname{cosec} A – \cot A)(\operatorname{cosec} A + \cot A)$$

$$= (\operatorname{cosec} A + \cot A)\,\big[1 – \operatorname{cosec} A + \cot A\big]$$

The bracket $1 – \operatorname{cosec} A + \cot A$ is exactly the denominator, so it cancels:

$$\text{LHS} = \frac{(\operatorname{cosec} A + \cot A)(\cot A + 1 – \operatorname{cosec} A)}{\cot A + 1 – \operatorname{cosec} A} = \operatorname{cosec} A + \cot A = \text{RHS}$$

Question 4 (vi)

$$\sqrt{\frac{1 + \sin A}{1 – \sin A}} = \sec A + \tan A$$

Solution. A square root of a fraction only simplifies if the fraction is a perfect square. Multiply top and bottom inside the root by $1 + \sin A$, which makes the numerator a square and turns the denominator into $1 – \sin^2 A = \cos^2 A$:

$$\text{LHS} = \sqrt{\frac{(1 + \sin A)^2}{(1 – \sin A)(1 + \sin A)}} = \sqrt{\frac{(1 + \sin A)^2}{\cos^2 A}} = \frac{1 + \sin A}{\cos A}$$

The positive root is the right one, because $1 + \sin A$ and $\cos A$ are both positive for an acute angle. Splitting the fraction,

$$\frac{1}{\cos A} + \frac{\sin A}{\cos A} = \sec A + \tan A = \text{RHS}$$

Question 4 (vii)

$$\frac{\sin\theta – 2\sin^3\theta}{2\cos^3\theta – \cos\theta} = \tan\theta$$

Solution. Take out $\sin\theta$ from the numerator and $\cos\theta$ from the denominator; the $\frac{\sin\theta}{\cos\theta}$ that appears is already the answer, so what is left must come to $1$:

$$\text{LHS} = \frac{\sin\theta\,(1 – 2\sin^2\theta)}{\cos\theta\,(2\cos^2\theta – 1)}$$

Rewrite the numerator’s bracket using $\sin^2\theta = 1 – \cos^2\theta$:

$$1 – 2\sin^2\theta = 1 – 2(1 – \cos^2\theta) = 2\cos^2\theta – 1$$

So the two brackets are equal, and they cancel (they are non-zero wherever the expression is defined; both vanish at $\theta = 45^\circ$):

$$\text{LHS} = \frac{\sin\theta}{\cos\theta} = \tan\theta = \text{RHS}$$

Question 4 (viii)

$$(\sin A + \operatorname{cosec} A)^2 + (\cos A + \sec A)^2 = 7 + \tan^2 A + \cot^2 A$$

Solution. Expand both squares. The cross terms are where the numbers come from, because $\sin A \operatorname{cosec} A = 1$ and $\cos A \sec A = 1$:

$$\text{LHS} = \sin^2 A + 2\sin A\operatorname{cosec} A + \operatorname{cosec}^2 A + \cos^2 A + 2\cos A\sec A + \sec^2 A$$

$$= (\sin^2 A + \cos^2 A) + 2 + 2 + \operatorname{cosec}^2 A + \sec^2 A$$

Replace each of the remaining squares by its identity:

$$= 1 + 4 + (1 + \cot^2 A) + (1 + \tan^2 A) = 7 + \tan^2 A + \cot^2 A = \text{RHS}$$

Question 4 (ix)

$$(\operatorname{cosec} A – \sin A)(\sec A – \cos A) = \frac{1}{\tan A + \cot A}$$

[Hint: Simplify LHS and RHS separately.]

Solution. Reduce each side to $\sin$ and $\cos$.

$$\text{LHS} = \left(\frac{1 – \sin^2 A}{\sin A}\right)\left(\frac{1 – \cos^2 A}{\cos A}\right) = \frac{\cos^2 A}{\sin A} \cdot \frac{\sin^2 A}{\cos A} = \sin A\cos A$$

$$\text{RHS} = \frac{1}{\dfrac{\sin A}{\cos A} + \dfrac{\cos A}{\sin A}} = \frac{1}{\dfrac{\sin^2 A + \cos^2 A}{\sin A\cos A}} = \frac{\sin A\cos A}{1} = \sin A\cos A$$

Both sides equal $\sin A\cos A$.

Question 4 (x)

$$\left(\frac{1 + \tan^2 A}{1 + \cot^2 A}\right) = \left(\frac{1 – \tan A}{1 – \cot A}\right)^2 = \tan^2 A$$

Solution. Two separate claims: each of the first two expressions equals $\tan^2 A$.

The first is question 3(iv) again:

$$\frac{1 + \tan^2 A}{1 + \cot^2 A} = \frac{\sec^2 A}{\operatorname{cosec}^2 A} = \frac{\sin^2 A}{\cos^2 A} = \tan^2 A$$

For the second, write $\cot A = \frac{1}{\tan A}$ and clear the small fraction:

$$\frac{1 – \tan A}{1 – \dfrac{1}{\tan A}} = \frac{1 – \tan A}{\dfrac{\tan A – 1}{\tan A}} = \frac{\tan A\,(1 – \tan A)}{\tan A – 1} = -\tan A$$

since $1 – \tan A$ and $\tan A – 1$ differ only in sign (and are non-zero, the expression being undefined at $A = 45^\circ$). Squaring removes the minus sign:

$$\left(\frac{1 – \tan A}{1 – \cot A}\right)^2 = (-\tan A)^2 = \tan^2 A$$

All three expressions are therefore equal.

Common mistakes

• Questions 1 and 2, writing $\sin A = \sqrt{1 + \cot^2 A}$. That is $\operatorname{cosec} A$. The identity $1 + \cot^2 A = \operatorname{cosec}^2 A$ delivers the reciprocal of sine, so one more step is needed. A quick check: at $A = 45^\circ$, $\sqrt{1 + 1} = \sqrt2$, which cannot be a sine.
• Question 3(ii), expanding $(\sin\theta + \cos\theta)^2$ as $1$. The middle term $2\sin\theta\cos\theta$ is the whole answer. Without it the product comes out as $0$ — which is option (A), placed there to catch exactly this slip.
• Question 4(iii), losing the minus sign. $1 – \tan\theta = \frac{\cos\theta – \sin\theta}{\cos\theta}$, the opposite way round from the other term’s $\sin\theta – \cos\theta$. Ignore the sign and the numerator becomes $\sin^3\theta + \cos^3\theta$, which does not factor against the denominator, and the proof stalls.
• Question 4(v), cross-multiplying and expanding. It can be made to work, but it is long, error-prone, and not what the question asks: it names the identity $\operatorname{cosec}^2 A = 1 + \cot^2 A$. Divide by $\sin A$ first, then replace the $1$.
• Question 4(vi), writing $\sqrt{1 – \sin^2 A}$ as $1 – \sin A$. A square root does not split across a difference. Multiplying by $1 + \sin A$ is what turns the inside into a perfect square.
• Question 4(vii), cancelling brackets that only look different. $1 – 2\sin^2\theta$ and $2\cos^2\theta – 1$ are equal, but that has to be shown — one line with $\sin^2\theta = 1 – \cos^2\theta$ — before they can cancel.
• Question 4(viii), dropping the cross terms. $2\sin A\operatorname{cosec} A$ and $2\cos A\sec A$ are $2$ each, and together they supply $4$ of the $7$. Squaring a sum as the sum of squares loses them.

Practise next

• Chapter 9, Exercise 9.1 — heights and distances, where the ratios are put to work on real lengths.
• Exercise 8.2 — revise the standard values; substituting $A = 30^\circ$ into any identity above is a fast way to check it before proving it.
• Class 11 Trigonometric Functions, Exercise 3.3 — the same kind of proof, now with compound and double angles.