Identities and Simplification

The Three Identities from One Right Triangle

Take a right triangle OAP whose hypotenuse OP is 1 unit long, with the angle $\theta$ at O. Then $\sin\theta = \frac{AP}{OP} = AP$ and $\cos\theta = \frac{OA}{OP} = OA$. So the two shorter sides are simply $\sin\theta$ and $\cos\theta$.

Pythagoras on this triangle gives the first and most basic identity:

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

Here $\sin^2\theta$ means $(\sin\theta)^2$, the square of the ratio. It does not mean $\sin(\theta^2)$.

The other two identities come from dividing this one. Divide every term by $\cos^2\theta$:

$$\frac{\sin^2\theta}{\cos^2\theta} + 1 = \frac{1}{\cos^2\theta} \Rightarrow \tan^2\theta + 1 = \sec^2\theta$$

Divide every term by $\sin^2\theta$ instead:

$$1 + \frac{\cos^2\theta}{\sin^2\theta} = \frac{1}{\sin^2\theta} \Rightarrow 1 + \cot^2\theta = \operatorname{cosec}^2\theta$$

The same thing in pictures: shrink or stretch the first triangle until its base is 1, and its sides become 1, $\tan\theta$ and $\sec\theta$; stretch it until its height is 1, and they become $\cot\theta$, 1 and $\operatorname{cosec}\theta$. Pythagoras on each triangle gives one identity.

A check with $\theta = 30^\circ$: $\sin^2 30^\circ + \cos^2 30^\circ = \frac{1}{4} + \frac{3}{4} = 1$, and $1 + \tan^2 30^\circ = 1 + \frac{1}{3} = \frac{4}{3}$, which is $\sec^2 30^\circ$ because $\sec 30^\circ = \frac{2}{\sqrt{3}}$.

These hold for every angle at which the ratios are defined. That is what makes them identities, and it is the difference between an identity and an equation such as $\sin\theta = \frac{1}{2}$, which is true only for particular angles.

Every Rearrangement, and the Factor Forms

In questions the identities seldom appear in the shape you learnt them. Know each one in all its forms:

Identity	Rearranged	Factor form
$\sin^2\theta + \cos^2\theta = 1$	$\sin^2\theta = 1 - \cos^2\theta$, $\cos^2\theta = 1 - \sin^2\theta$	$(1 - \sin\theta)(1 + \sin\theta) = \cos^2\theta$, $(1 - \cos\theta)(1 + \cos\theta) = \sin^2\theta$
$\sec^2\theta - \tan^2\theta = 1$	$\sec^2\theta = 1 + \tan^2\theta$, $\tan^2\theta = \sec^2\theta - 1$	$(\sec\theta - 1)(\sec\theta + 1) = \tan^2\theta$
$\operatorname{cosec}^2\theta - \cot^2\theta = 1$	$\operatorname{cosec}^2\theta = 1 + \cot^2\theta$, $\cot^2\theta = \operatorname{cosec}^2\theta - 1$	$(\operatorname{cosec}\theta - 1)(\operatorname{cosec}\theta + 1) = \cot^2\theta$

When you see 1 minus a square, or two brackets that differ only in a sign, look for one of these. A few lines show how fast they work:

• $(1 - \cos^2\theta)\operatorname{cosec}^2\theta = \sin^2\theta \times \frac{1}{\sin^2\theta} = 1$
• $\cos^2\theta(1 + \tan^2\theta) = \cos^2\theta \times \sec^2\theta = 1$
• $(\sec^2\theta - 1)(1 - \sin^2\theta) = \tan^2\theta\cos^2\theta = \sin^2\theta$
• $\frac{1}{1 + \tan^2\theta} + \frac{1}{1 + \cot^2\theta} = \cos^2\theta + \sin^2\theta = 1$

The graph shows $\sin^2\theta$ and $\cos^2\theta$ for angles from 0° to 90°. Whatever the angle, the two heights add up to 1. They are equal only at 45°, where each is $\frac{1}{2}$; that fact matters later, when options are tested at 45°.

Numbers work the same way. $5\sin^2 37^\circ + 5\cos^2 37^\circ = 5$ and $7\sec^2 20^\circ - 7\tan^2 20^\circ = 7$, without knowing any of the separate values. The angle can be anything, as long as it is the same angle in both terms.

Same angle only. $\sin^2 30^\circ + \cos^2 60^\circ$ is not 1. It is $\frac{1}{4} + \frac{1}{4} = \frac{1}{2}$. The identity joins $\sin^2$ and $\cos^2$ of one angle; for two different angles it says nothing (unless they are complementary, which is a separate rule, see Standard Angles and Complementary Angles).

Square of a sum is not a sum of squares. $\sin^2\theta + \cos^2\theta = 1$ does not give $\sin\theta + \cos\theta = 1$. At $\theta = 45^\circ$ the sum $\sin\theta + \cos\theta$ is $\sqrt{2}$.

When Stuck, Write Everything in $\sin\theta$ and $\cos\theta$

Most simplifications become routine once tan, cot, sec and cosec are replaced by sin and cos. In rough work write s for $\sin\theta$ and c for $\cos\theta$ to save time, and look for $s^2 + c^2$ to turn into 1.

• $\tan\theta + \cot\theta = \frac{s}{c} + \frac{c}{s} = \frac{s^2 + c^2}{sc} = \frac{1}{\sin\theta\cos\theta} = \sec\theta\operatorname{cosec}\theta$
• $\sec\theta - \cos\theta = \frac{1}{c} - c = \frac{1 - c^2}{c} = \frac{s^2}{c} = \sin\theta\tan\theta$
• $\operatorname{cosec}\theta - \sin\theta = \frac{1 - s^2}{s} = \frac{c^2}{s} = \cos\theta\cot\theta$
• $\sec^2\theta + \operatorname{cosec}^2\theta = \frac{1}{c^2} + \frac{1}{s^2} = \frac{s^2 + c^2}{s^2c^2} = \sec^2\theta\operatorname{cosec}^2\theta$

The last line says that the sum and the product of $\sec^2\theta$ and $\operatorname{cosec}^2\theta$ are equal. It looks odd, and it is asked on its own.

Combine two of them. From the second and third lines,

$$(\sec\theta - \cos\theta)(\operatorname{cosec}\theta - \sin\theta) = \frac{s^2}{c} \times \frac{c^2}{s} = sc$$

and since $\tan\theta + \cot\theta = \frac{1}{sc}$,

$$(\sec\theta - \cos\theta)(\operatorname{cosec}\theta - \sin\theta)(\tan\theta + \cot\theta) = 1$$

Grouping helps. $\sin\theta(1 + \tan\theta) + \cos\theta(1 + \cot\theta) = s + \frac{s^2}{c} + c + \frac{c^2}{s}$. Pair each term with the one that has the other letter at the bottom:

$$\left(s + \frac{c^2}{s}\right) + \left(c + \frac{s^2}{c}\right) = \frac{s^2 + c^2}{s} + \frac{c^2 + s^2}{c} = \frac{1}{s} + \frac{1}{c}$$

So the expression is $\operatorname{cosec}\theta + \sec\theta$.

Use this method when no identity jumps out at you. It always works, though it is not always the shortest road; with practice you will see the shorter routes in the next few topics.

The Reciprocal Pairs: $\sec\theta \pm \tan\theta$ and $\operatorname{cosec}\theta \pm \cot\theta$

Factorise $\sec^2\theta - \tan^2\theta = 1$ as a difference of squares:

$$(\sec\theta + \tan\theta)(\sec\theta - \tan\theta) = 1$$

So $\sec\theta + \tan\theta$ and $\sec\theta - \tan\theta$ are reciprocals of each other. In the same way

$$(\operatorname{cosec}\theta + \cot\theta)(\operatorname{cosec}\theta - \cot\theta) = 1$$

Written in sin and cos the four expressions look quite different, and questions use both forms:

Expression	In sin and cos	Other form
$\sec\theta + \tan\theta$	$\frac{1 + \sin\theta}{\cos\theta}$	$\frac{\cos\theta}{1 - \sin\theta}$
$\sec\theta - \tan\theta$	$\frac{1 - \sin\theta}{\cos\theta}$	$\frac{\cos\theta}{1 + \sin\theta}$
$\operatorname{cosec}\theta + \cot\theta$	$\frac{1 + \cos\theta}{\sin\theta}$	$\frac{\sin\theta}{1 - \cos\theta}$
$\operatorname{cosec}\theta - \cot\theta$	$\frac{1 - \cos\theta}{\sin\theta}$	$\frac{\sin\theta}{1 + \cos\theta}$

The last column comes from multiplying top and bottom by the partner bracket. For example $\frac{1 - \sin\theta}{\cos\theta} \times \frac{1 + \sin\theta}{1 + \sin\theta} = \frac{\cos^2\theta}{\cos\theta(1 + \sin\theta)} = \frac{\cos\theta}{1 + \sin\theta}$.

Using the pairs.

• $\frac{1}{\sec\theta - \tan\theta} = \sec\theta + \tan\theta$, so $\frac{1}{\sec\theta - \tan\theta} - \tan\theta = \sec\theta$.
• $\frac{1}{\operatorname{cosec}\theta - \cot\theta} - \frac{1}{\sin\theta} = (\operatorname{cosec}\theta + \cot\theta) - \operatorname{cosec}\theta = \cot\theta$.
• $\frac{\cos\theta}{1 - \sin\theta}$ is $\sec\theta + \tan\theta$ straight from the table.

When one member of a pair is given as a number, the other is its reciprocal: $\sec\theta + \tan\theta = 4$ means $\sec\theta - \tan\theta = \frac{1}{4}$. Finding $\sec\theta$ and $\sin\theta$ from there is done in Values from a Condition, Maximum and Minimum.

[SSC Tip] The reciprocal of a sum is not the sum of the reciprocals. $\frac{1}{\sec\theta + \tan\theta}$ is $\sec\theta - \tan\theta$; writing it as $\cos\theta + \cot\theta$ adds the reciprocals separately, and that form can appear among the options.

Square Roots: Make the Bottom a Perfect Square

$\sqrt{\frac{1 - \sin\theta}{1 + \sin\theta}}$ cannot be simplified as it stands. Multiply the top and the bottom inside the root by $1 - \sin\theta$, so that the bottom becomes $1 - \sin^2\theta = \cos^2\theta$:

$$\sqrt{\frac{1 - \sin\theta}{1 + \sin\theta}} = \sqrt{\frac{(1 - \sin\theta)^2}{1 - \sin^2\theta}} = \sqrt{\frac{(1 - \sin\theta)^2}{\cos^2\theta}} = \frac{1 - \sin\theta}{\cos\theta} = \sec\theta - \tan\theta$$

For an acute angle both $1 - \sin\theta$ and $\cos\theta$ are positive, so the square root comes off cleanly. SSC questions take $\theta$ acute unless they say otherwise.

Check at $\theta = 30^\circ$: inside the root is $\frac{1/2}{3/2} = \frac{1}{3}$, so the left side is $\frac{1}{\sqrt{3}}$; the right side is $\frac{2}{\sqrt{3}} - \frac{1}{\sqrt{3}} = \frac{1}{\sqrt{3}}$.

The whole family is done the same way: multiply by whatever turns the bottom into a square.

Expression ($\theta$ acute)	Equals
$\sqrt{\frac{1 - \sin\theta}{1 + \sin\theta}}$	$\sec\theta - \tan\theta$
$\sqrt{\frac{1 + \sin\theta}{1 - \sin\theta}}$	$\sec\theta + \tan\theta$
$\sqrt{\frac{1 - \cos\theta}{1 + \cos\theta}}$	$\operatorname{cosec}\theta - \cot\theta$
$\sqrt{\frac{1 + \cos\theta}{1 - \cos\theta}}$	$\operatorname{cosec}\theta + \cot\theta$
$\sqrt{\frac{\sec\theta - 1}{\sec\theta + 1}}$	$\operatorname{cosec}\theta - \cot\theta$
$\sqrt{\frac{\operatorname{cosec}\theta - 1}{\operatorname{cosec}\theta + 1}}$	$\sec\theta - \tan\theta$

Look carefully at the last two rows. A root with sec inside gives cosec and cot outside, and the other way round. Multiply the top and bottom inside by $\cos\theta$ and $\frac{\sec\theta - 1}{\sec\theta + 1}$ turns into $\frac{1 - \cos\theta}{1 + \cos\theta}$, the third row.

Two roots together. A root and its upside-down partner add or subtract neatly:

$$\sqrt{\frac{1 - \sin\theta}{1 + \sin\theta}} + \sqrt{\frac{1 + \sin\theta}{1 - \sin\theta}} = (\sec\theta - \tan\theta) + (\sec\theta + \tan\theta) = 2\sec\theta$$

and the second minus the first is $2\tan\theta$. The cos pair gives $2\operatorname{cosec}\theta$ and $2\cot\theta$ in the same way.

Adding Two Fractions: Cross-Multiply, Then Look for 1

Take $\frac{\sin\theta}{1 + \cos\theta} + \frac{1 + \cos\theta}{\sin\theta}$. Over the common denominator $\sin\theta(1 + \cos\theta)$, the top is

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

The bracket $(1 + \cos\theta)$ cancels with the bottom and $\frac{2}{\sin\theta} = 2\operatorname{cosec}\theta$ is left. The cos copy, $\frac{\cos\theta}{1 + \sin\theta} + \frac{1 + \sin\theta}{\cos\theta}$, comes to $2\sec\theta$ by the same steps.

Bottoms $1 + x$ and $1 - x$. The common denominator is $1 - x^2$, which one of the identities turns into a single square.

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

A harder pair in tan and cot. $\frac{\tan\theta}{1 - \cot\theta} + \frac{\cot\theta}{1 - \tan\theta}$. Write t for $\tan\theta$, so $\cot\theta = \frac{1}{t}$. The first term is $\frac{t}{1 - \frac{1}{t}} = \frac{t^2}{t - 1}$ and the second is $\frac{1}{t(1 - t)} = -\frac{1}{t(t - 1)}$. Together

$$\frac{t^3 - 1}{t(t - 1)} = \frac{t^2 + t + 1}{t} = t + 1 + \frac{1}{t}$$

using $t^3 - 1 = (t - 1)(t^2 + t + 1)$ from Chapter 11. So the value is $1 + \tan\theta + \cot\theta = 1 + \sec\theta\operatorname{cosec}\theta$.

When the bottoms have tan or cot, it is often quicker to turn everything into sin and cos first: $\frac{\cos\theta}{1 - \tan\theta}$ becomes $\frac{\cos^2\theta}{\cos\theta - \sin\theta}$ after multiplying top and bottom by $\cos\theta$.

Three-Term Fractions: Write the 1 as $\sec^2\theta - \tan^2\theta$

$\frac{\tan\theta + \sec\theta - 1}{\tan\theta - \sec\theta + 1}$ comes up often in CGL papers, and cross-multiplying gets nowhere. The way in is to replace the 1 on top by $\sec^2\theta - \tan^2\theta$:

$$\tan\theta + \sec\theta - (\sec^2\theta - \tan^2\theta) = (\sec\theta + \tan\theta) - (\sec\theta + \tan\theta)(\sec\theta - \tan\theta)$$

$$= (\sec\theta + \tan\theta)(1 - \sec\theta + \tan\theta)$$

The second bracket, $1 - \sec\theta + \tan\theta$, is exactly the bottom. It cancels:

$$\frac{\tan\theta + \sec\theta - 1}{\tan\theta - \sec\theta + 1} = \sec\theta + \tan\theta = \frac{1 + \sin\theta}{\cos\theta}$$

At $\theta = 60^\circ$ ($\tan\theta = \sqrt{3}$, $\sec\theta = 2$) the fraction is $\frac{\sqrt{3} + 1}{\sqrt{3} - 1} = 2 + \sqrt{3}$, which is $\sec 60^\circ + \tan 60^\circ$.

The cot and cosec copy. Writing 1 as $\operatorname{cosec}^2\theta - \cot^2\theta$,

$$\frac{\cot\theta + \operatorname{cosec}\theta - 1}{\cot\theta - \operatorname{cosec}\theta + 1} = \operatorname{cosec}\theta + \cot\theta = \frac{1 + \cos\theta}{\sin\theta}$$

The same fraction in disguise. $\frac{\sin\theta - \cos\theta + 1}{\sin\theta + \cos\theta - 1}$: divide every term by $\cos\theta$ and it becomes $\frac{\tan\theta - 1 + \sec\theta}{\tan\theta + 1 - \sec\theta}$, the first fraction above. So it also equals $\sec\theta + \tan\theta$. Dividing by $\sin\theta$ instead shows that $\frac{\cos\theta - \sin\theta + 1}{\cos\theta + \sin\theta - 1} = \operatorname{cosec}\theta + \cot\theta$.

When the 1 sits in front. $\frac{1 + \sec\theta - \tan\theta}{1 + \sec\theta + \tan\theta}$: write the top as $(\sec^2\theta - \tan^2\theta) + (\sec\theta - \tan\theta) = (\sec\theta - \tan\theta)(\sec\theta + \tan\theta + 1)$. The second bracket is the bottom, so the fraction is $\sec\theta - \tan\theta = \frac{1 - \sin\theta}{\cos\theta}$.

In each case, choose the factorisation that produces a copy of the bottom; the rest is cancelling.

Products and Squares That Come Out as a Number

Some long expressions collapse to a plain number, or to something much shorter. Write them in sin and cos and expand.

$(1 + \tan\theta + \sec\theta)(1 + \cot\theta - \operatorname{cosec}\theta)$. In sin and cos the two brackets are

$$\frac{\cos\theta + \sin\theta + 1}{\cos\theta} \quad \text{and} \quad \frac{\sin\theta + \cos\theta - 1}{\sin\theta}$$

With $a = \sin\theta + \cos\theta$ the tops are $a + 1$ and $a - 1$, whose product is

$$a^2 - 1 = (\sin\theta + \cos\theta)^2 - 1 = 2\sin\theta\cos\theta$$

The bottoms multiply to $\sin\theta\cos\theta$, so the product is 2, whatever $\theta$ is. At 45° you can see it directly: $(2 + \sqrt{2})(2 - \sqrt{2}) = 4 - 2 = 2$.

$(\sin\theta + \operatorname{cosec}\theta)^2 + (\cos\theta + \sec\theta)^2$. Since $\sin\theta\operatorname{cosec}\theta = 1$, the first square is $\sin^2\theta + 2 + \operatorname{cosec}^2\theta$, and the second is $\cos^2\theta + 2 + \sec^2\theta$. Adding,

$$1 + 4 + \operatorname{cosec}^2\theta + \sec^2\theta = 5 + (1 + \cot^2\theta) + (1 + \tan^2\theta) = 7 + \tan^2\theta + \cot^2\theta$$

The answer is not a plain number here, and it is easy to stop at 5 or to drop the two 1s that come with $\operatorname{cosec}^2\theta$ and $\sec^2\theta$.

$(\sec\theta + \tan\theta - 1)(\sec\theta - \tan\theta + 1)$. Group as $[\sec\theta + (\tan\theta - 1)][\sec\theta - (\tan\theta - 1)]$, a difference of squares:

$$\sec^2\theta - (\tan\theta - 1)^2 = \sec^2\theta - \tan^2\theta + 2\tan\theta - 1 = 2\tan\theta$$

Quick ones. $\frac{\sin\theta}{\operatorname{cosec}\theta} + \frac{\cos\theta}{\sec\theta} = \sin^2\theta + \cos^2\theta = 1$. And $(\sec\theta - \cos\theta)(\operatorname{cosec}\theta - \sin\theta)(\tan\theta + \cot\theta) = 1$, worked out earlier in these notes.

When the options are all numbers, the expression must be a constant, so any convenient angle gives the answer. That check is set out in the part on testing options below.

Fourth and Sixth Powers

Treat $\sin^2\theta$ and $\cos^2\theta$ as two numbers a and b with a + b = 1, and use the algebra identities of Chapter 11.

Difference of fourth powers. $a^2 - b^2 = (a - b)(a + b)$, and the second bracket is 1:

$$\sin^4\theta - \cos^4\theta = (\sin^2\theta - \cos^2\theta)(\sin^2\theta + \cos^2\theta) = \sin^2\theta - \cos^2\theta$$

This can also be written $1 - 2\cos^2\theta$ or $2\sin^2\theta - 1$, and the options may use any of the three forms. The sec and tan copy is $\sec^4\theta - \tan^4\theta = \sec^2\theta + \tan^2\theta = 1 + 2\tan^2\theta$.

Sum of fourth powers. $a^2 + b^2 = (a + b)^2 - 2ab$:

$$\sin^4\theta + \cos^4\theta = 1 - 2\sin^2\theta\cos^2\theta$$

Sum of sixth powers. $a^3 + b^3 = (a + b)^3 - 3ab(a + b)$:

$$\sin^6\theta + \cos^6\theta = 1 - 3\sin^2\theta\cos^2\theta$$

At $\theta = 45^\circ$, $\sin^2\theta\cos^2\theta = \frac{1}{4}$, so $\sin^4\theta + \cos^4\theta = \frac{1}{2}$ and $\sin^6\theta + \cos^6\theta = \frac{1}{4}$. Direct working agrees: $2 \times \left(\frac{1}{\sqrt{2}}\right)^4 = \frac{1}{2}$.

Combinations that cancel. Both formulas contain the same product. Call it $x = \sin^2\theta\cos^2\theta$; then

$$2(\sin^6\theta + \cos^6\theta) - 3(\sin^4\theta + \cos^4\theta) = 2 - 6x - 3 + 6x = -1$$

so $2(\sin^6\theta + \cos^6\theta) - 3(\sin^4\theta + \cos^4\theta) + 1 = 0$ for every $\theta$.

A longer one in the same spirit uses $y = \sin\theta\cos\theta$, with $(\sin\theta - \cos\theta)^2 = 1 - 2y$ and $(\sin\theta + \cos\theta)^2 = 1 + 2y$:

$$3(\sin\theta - \cos\theta)^4 + 6(\sin\theta + \cos\theta)^2 + 4(\sin^6\theta + \cos^6\theta) = 3(1 - 2y)^2 + 6(1 + 2y) + 4(1 - 3y^2) = 13$$

[SSC Tip] When the options are plain numbers, put $\theta = 0^\circ$ ($\sin\theta = 0$, $\cos\theta = 1$). The long expression above becomes 3 + 6 + 4 = 13 in one line. Use 0° only if nothing in the question becomes undefined there, as $\tan\theta$ and $\sec\theta$ are fine at 0° but $\cot\theta$ and $\operatorname{cosec}\theta$ are not.

Testing the Options by Putting In 45°, 30° or 60°

When the options are expressions in $\theta$, the right option equals the question at every angle. So put in one angle, find the value of the question, and keep only the options that give the same value.

Take $\frac{\sin\theta}{1 + \cos\theta} + \frac{1 + \cos\theta}{\sin\theta}$ with the options $2\sec\theta$, $2\operatorname{cosec}\theta$, $2\tan\theta$ and $2\cot\theta$.

• At 45°: $\frac{\sin 45^\circ}{1 + \cos 45^\circ} = \frac{1}{\sqrt{2} + 1} = \sqrt{2} - 1$ and its upside-down partner is $\sqrt{2} + 1$, so the question is $2\sqrt{2}$. The options give $2\sqrt{2}$, $2\sqrt{2}$, 2 and 2. Two options drop out, but two survive.
• At 60°: the question is $\frac{\sqrt{3}/2}{3/2} + \frac{3/2}{\sqrt{3}/2} = \frac{1}{\sqrt{3}} + \sqrt{3} = \frac{4}{\sqrt{3}}$. Now $2\sec 60^\circ = 4$ while $2\operatorname{cosec} 60^\circ = \frac{4}{\sqrt{3}}$. The answer is $2\operatorname{cosec}\theta$.

When the check fails or misleads.

• At 45°, $\sin\theta = \cos\theta$, $\tan\theta = \cot\theta$ and $\sec\theta = \operatorname{cosec}\theta$. If two options differ only by such a swap, 45° cannot separate them. Use 30° or 60°.
• The question may be undefined at the angle you pick. $\frac{\tan\theta}{1 - \cot\theta} + \frac{\cot\theta}{1 - \tan\theta}$ has $1 - \tan 45^\circ = 0$ at the bottom, so 45° tells you nothing there. Avoid 0° and 90° as well when tan, sec, cot or cosec appear.
• Two different expressions can agree at one angle by chance. If more than one option survives, test a second angle before choosing.
• Keep the values exact ($\frac{2}{\sqrt{3}}$, not 1.15). Rounded decimals can make two different options look equal.

Order of work. Try 45° first, since its arithmetic is lightest. If two options are left, or the question breaks at 45°, switch to 60° or 30°. If the options are plain numbers, $\theta = 0^\circ$ is often lighter still.

The check is a tool for multiple-choice questions only. In a question that asks you to prove or simplify fully, the algebra of the earlier parts is still needed, and the check simply confirms your final line.

Eliminating $\theta$

Two or three equations contain $\theta$, and the question asks for a relation between the other letters with no $\theta$ in it. Make each ratio the subject, then use an identity those ratios satisfy.

$x = a\sec\theta$, $y = b\tan\theta$. Then $\sec\theta = \frac{x}{a}$ and $\tan\theta = \frac{y}{b}$, and $\sec^2\theta - \tan^2\theta = 1$ gives

$$\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$$

With numbers, $x = 2\sec\theta$ and $y = 5\tan\theta$ give $\frac{x^2}{4} - \frac{y^2}{25} = 1$, that is $25x^2 - 4y^2 = 100$.

$x = a\cos\theta$, $y = b\sin\theta$ gives $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$, this time from $\cos^2\theta + \sin^2\theta = 1$. With cosec and cot the identity is $\operatorname{cosec}^2\theta - \cot^2\theta = 1$.

Three letters. $x = r\cos t$, $y = r\sin t\cos u$, $z = r\sin t\sin u$. Here y and z share $r\sin t$, so add their squares first:

$$y^2 + z^2 = r^2\sin^2 t(\cos^2 u + \sin^2 u) = r^2\sin^2 t$$

Now add $x^2 = r^2\cos^2 t$:

$$x^2 + y^2 + z^2 = r^2(\sin^2 t + \cos^2 t) = r^2$$

The identity is used twice, once for each angle. Pair the two equations that share a factor first.

Sum and difference given. If $x = \sec\theta + \tan\theta$ and $y = \sec\theta - \tan\theta$, then xy = 1 straight from the reciprocal pair. The cosec and cot pair works the same way.

[Tier 2] Two longer eliminations.

• $\tan\theta + \sin\theta = m$ and $\tan\theta - \sin\theta = n$. Then $m^2 - n^2 = (m + n)(m - n) = 4\tan\theta\sin\theta$, while $mn = \tan^2\theta - \sin^2\theta = \sin^2\theta(\sec^2\theta - 1) = \tan^2\theta\sin^2\theta$. So $m^2 - n^2 = 4\sqrt{mn}$.
• $x = \operatorname{cosec}\theta - \sin\theta$ and $y = \sec\theta - \cos\theta$. In sin and cos, $x = \frac{\cos^2\theta}{\sin\theta}$ and $y = \frac{\sin^2\theta}{\cos\theta}$. So $x^2y = \cos^3\theta$ and $xy^2 = \sin^3\theta$, which gives $(x^2y)^{\frac{2}{3}} + (xy^2)^{\frac{2}{3}} = 1$.

Solved Examples

Question 1

What is $(1 + \cot^2\theta)(1 - \cos\theta)(1 + \cos\theta)$ equal to? (a) 0 (b) 1 (c) $\operatorname{cosec}^2\theta$ (d) $\sin^2\theta$

Answer. The last two brackets multiply to $1 - \cos^2\theta = \sin^2\theta$, and the first bracket is $\operatorname{cosec}^2\theta$. So the whole thing is $\operatorname{cosec}^2\theta\sin^2\theta$, which is (b) 1.

Question 2

$\frac{\sin^2\theta}{1 - \cos\theta}$ is equal to (a) $1 + \cos\theta$ (b) $1 - \cos\theta$ (c) $1 + \sin\theta$ (d) $\cos\theta$

Answer. I write the top as $1 - \cos^2\theta = (1 - \cos\theta)(1 + \cos\theta)$. One bracket cancels with the bottom, and I am left with (a) $1 + \cos\theta$. A check at 60°: $\frac{3}{4} \div \frac{1}{2} = \frac{3}{2}$, and $1 + \cos 60^\circ = \frac{3}{2}$ too.

Question 3

$\sec^4\theta - \sec^2\theta$ is equal to (a) $\tan^4\theta - \tan^2\theta$ (b) $1 + \tan^4\theta$ (c) $\tan^2\theta$ (d) $\tan^4\theta + \tan^2\theta$

Answer. I take $\sec^2\theta$ common: $\sec^2\theta(\sec^2\theta - 1)$. The bracket is $\tan^2\theta$ and the outside factor is $1 + \tan^2\theta$, so I get $(1 + \tan^2\theta)\tan^2\theta$. Multiplying out gives (d) $\tan^4\theta + \tan^2\theta$.

Question 4

$\sec\theta\operatorname{cosec}\theta - \cot\theta$ is equal to (a) $\tan\theta$ (b) $\cot\theta$ (c) $\sin\theta$ (d) $\sec\theta$

Answer. In sin and cos: $\frac{1}{\sin\theta\cos\theta} - \frac{\cos\theta}{\sin\theta} = \frac{1 - \cos^2\theta}{\sin\theta\cos\theta} = \frac{\sin^2\theta}{\sin\theta\cos\theta}$. One $\sin\theta$ cancels and $\frac{\sin\theta}{\cos\theta}$ is left, which is (a) $\tan\theta$.

Question 5

What is the value of $(\sec\theta - \cos\theta)(\operatorname{cosec}\theta - \sin\theta)(\tan\theta + \cot\theta)$? (a) 0 (b) 2 (c) 1 (d) $\sin\theta\cos\theta$

Answer. I take the brackets one at a time. $\sec\theta - \cos\theta = \frac{1 - \cos^2\theta}{\cos\theta} = \frac{\sin^2\theta}{\cos\theta}$, $\operatorname{cosec}\theta - \sin\theta = \frac{\cos^2\theta}{\sin\theta}$ and $\tan\theta + \cot\theta = \frac{1}{\sin\theta\cos\theta}$. The first two multiply to $\sin\theta\cos\theta$, and the third cancels it, so the value is (c) 1. Stopping after two brackets leaves $\sin\theta\cos\theta$, which is only part of the product.

Question 6

$\frac{1}{\sec\theta - \tan\theta} - \frac{1}{\cos\theta}$ is equal to (a) $\tan\theta$ (b) $\sec\theta$ (c) $2\tan\theta$ (d) $\cot\theta$

Answer. $\sec\theta - \tan\theta$ and $\sec\theta + \tan\theta$ multiply to 1, so the first fraction is just $\sec\theta + \tan\theta$. The second fraction is $\sec\theta$. Subtracting, I am left with (a) $\tan\theta$.

Question 7

$\frac{\cos\theta}{1 - \sin\theta} - \tan\theta$ is equal to (a) $\cos\theta$ (b) $2\tan\theta$ (c) $\sec\theta$ (d) 1

Answer. I multiply the top and bottom of the first fraction by $1 + \sin\theta$: $\frac{\cos\theta(1 + \sin\theta)}{1 - \sin^2\theta} = \frac{\cos\theta(1 + \sin\theta)}{\cos^2\theta} = \frac{1 + \sin\theta}{\cos\theta}$. That is $\sec\theta + \tan\theta$, and taking away $\tan\theta$ gives (c) $\sec\theta$.

Question 8

For an acute angle $\theta$, $\sqrt{\frac{1 + \sin\theta}{1 - \sin\theta}} - \tan\theta$ is equal to (a) $\cos\theta$ (b) $\sec\theta$ (c) $\sec\theta - 2\tan\theta$ (d) 1

Answer. To make the bottom a square I multiply inside the root by $1 + \sin\theta$: $\sqrt{\frac{(1 + \sin\theta)^2}{1 - \sin^2\theta}} = \frac{1 + \sin\theta}{\cos\theta} = \sec\theta + \tan\theta$. Taking away $\tan\theta$ leaves (b) $\sec\theta$. Check at 30°: inside the root is $\frac{1 + \frac{1}{2}}{1 - \frac{1}{2}} = 3$, so the expression is $\sqrt{3} - \frac{1}{\sqrt{3}} = \frac{2}{\sqrt{3}}$, which is $\sec 30^\circ$.

Question 9

For an acute angle $\theta$, $\sqrt{\frac{1 - \cos\theta}{1 + \cos\theta}} + \sqrt{\frac{1 + \cos\theta}{1 - \cos\theta}}$ is equal to (a) $2\sec\theta$ (b) $2\cot\theta$ (c) $2\operatorname{cosec}\theta$ (d) 2

Answer. Making each bottom a square ($1 - \cos^2\theta = \sin^2\theta$), the first root is $\frac{1 - \cos\theta}{\sin\theta} = \operatorname{cosec}\theta - \cot\theta$ and the second is $\frac{1 + \cos\theta}{\sin\theta} = \operatorname{cosec}\theta + \cot\theta$. The cot terms cancel and (c) $2\operatorname{cosec}\theta$ is left. Putting $\theta = 45^\circ$ would not have settled this one, because $2\sec 45^\circ$ and $2\operatorname{cosec} 45^\circ$ are both $2\sqrt{2}$.

Question 10

For an acute angle $\theta$, $\sqrt{\frac{\sec\theta - 1}{\sec\theta + 1}}$ is equal to (a) $\sec\theta - \tan\theta$ (b) $\operatorname{cosec}\theta + \cot\theta$ (c) $\sec\theta + \tan\theta$ (d) $\operatorname{cosec}\theta - \cot\theta$

Answer. First I multiply the top and bottom inside by $\cos\theta$: $\frac{\sec\theta - 1}{\sec\theta + 1} = \frac{1 - \cos\theta}{1 + \cos\theta}$. Now it is the cos root, and multiplying inside by $1 - \cos\theta$ gives $\frac{1 - \cos\theta}{\sin\theta}$, which is (d) $\operatorname{cosec}\theta - \cot\theta$. A sec inside the root gives cosec and cot outside, which is easy to get backwards.

Question 11

$\frac{\cot\theta}{\operatorname{cosec}\theta - 1} + \frac{\cot\theta}{\operatorname{cosec}\theta + 1}$ is equal to (a) $2\operatorname{cosec}\theta$ (b) $2\sec\theta$ (c) $2\tan\theta$ (d) $2\cot\theta$

Answer. I test the options. At 45°, $\cot\theta = 1$ and $\operatorname{cosec}\theta = \sqrt{2}$, so the question is $\frac{1}{\sqrt{2} - 1} + \frac{1}{\sqrt{2} + 1} = (\sqrt{2} + 1) + (\sqrt{2} - 1) = 2\sqrt{2}$. That rules out $2\tan\theta$ and $2\cot\theta$ (both 2), but $2\sec\theta$ and $2\operatorname{cosec}\theta$ are both $2\sqrt{2}$ there. So I try 60°, where $\cot\theta = \frac{1}{\sqrt{3}}$ and $\operatorname{cosec}\theta = \frac{2}{\sqrt{3}}$. Multiplying top and bottom of each fraction by $\sqrt{3}$, the question is $\frac{1}{2 - \sqrt{3}} + \frac{1}{2 + \sqrt{3}} = (2 + \sqrt{3}) + (2 - \sqrt{3}) = 4$. Now $2\sec 60^\circ = 4$ but $2\operatorname{cosec} 60^\circ = \frac{4}{\sqrt{3}}$, so the answer is (b) $2\sec\theta$.

Question 12

$\frac{\tan\theta}{1 - \cot\theta} + \frac{\cot\theta}{1 - \tan\theta}$ is equal to (a) $\sec\theta\operatorname{cosec}\theta$ (b) $1 + \sec\theta\operatorname{cosec}\theta$ (c) $\sec\theta\operatorname{cosec}\theta - 1$ (d) $1 + \sin\theta\cos\theta$

Answer. 45° is no use here, because $1 - \tan 45^\circ = 0$. So I simplify, writing t for $\tan\theta$. The first term is $\frac{t}{1 - \frac{1}{t}} = \frac{t^2}{t - 1}$ and the second is $\frac{1}{t(1 - t)} = -\frac{1}{t(t - 1)}$. Together they make $\frac{t^3 - 1}{t(t - 1)} = \frac{t^2 + t + 1}{t} = t + 1 + \frac{1}{t}$. That is $1 + \tan\theta + \cot\theta$, and since $\tan\theta + \cot\theta = \sec\theta\operatorname{cosec}\theta$, the answer is (b) $1 + \sec\theta\operatorname{cosec}\theta$.