Trigonometric Ratios, Allied Angles and Identities | NIMCET

Quick answer: The six trigonometric ratios of an angle θ come from a point $(x, y)$ on its terminal side at distance $r$ from the origin: $\sin\theta = \frac{y}{r}$, $\cos\theta = \frac{x}{r}$, $\tan\theta = \frac{y}{x}$ and their reciprocals. The signs in each quadrant follow the ASTC pattern (All, Sin, Tan, Cos positive in quadrants I–IV). Any angle $n\cdot 90^\circ \pm \theta$ can be reduced to an acute angle, and the three Pythagorean identities, such as $\sec^2\theta - \tan^2\theta = 1$, handle most simplification questions.

Almost every other trigonometry topic in NIMCET — compound angles, equations, inverse functions, triangles — rests on the material on this page, so it is worth getting it completely secure.

Measuring angles: degrees and radians

One radian is the angle subtended at the centre of a circle by an arc equal in length to the radius. A full turn is $2\pi$ radians, so:

$\pi \text{ radians} = 180^\circ$

For a circle of radius $r$ and central angle θ in radians:

$l = r\theta, \qquad \text{Area of sector} = \frac{1}{2}r^2\theta$

These formulas are wrong if θ is in degrees — a frequent source of lost marks.

The six ratios and quadrant signs

Take a point $P(x, y)$ on the terminal side of θ, with $r = \sqrt{x^2 + y^2} > 0$.

Because $r$ is always positive, the sign of each ratio depends only on the signs of $x$ and $y$:

Ranges worth remembering

A quick consequence: for any non-zero real $x$, $\left|x + \frac{1}{x}\right| \ge 2$, so an equation like $\sin\theta = x + \frac{1}{x}$ has no real solution, because $\sin\theta$ can never exceed 1 in size. In contrast, $\sec\theta = x + \frac{1}{x}$ is always possible. Range arguments like this often eliminate options instantly.

Allied angles: reducing any angle

Angles of the form $n \cdot 90^\circ \pm \theta$ (with θ treated as acute) are called allied angles. Reduce them in three steps:

Step 1: Find the quadrant of $n \cdot 90^\circ \pm \theta$, assuming θ is acute.

Step 2: Fix the sign using ASTC for the original function in that quadrant.

Step 3: Decide the function.

• $n$ even ($180^\circ \pm \theta$, $360^\circ \pm \theta$): the function stays the same.
• $n$ odd ($90^\circ \pm \theta$, $270^\circ \pm \theta$): switch to the co-function — sin ↔ cos, tan ↔ cot, sec ↔ cosec.

Negative angles: cos and sec are even functions, so $\cos(-\theta) = \cos\theta$; sin, tan, cot and cosec are odd, so $\sin(-\theta) = -\sin\theta$.

Example: $\sin 1230^\circ$. Remove full turns: $1230^\circ - 3 \times 360^\circ = 150^\circ$, and $\sin 150^\circ = \sin(180^\circ - 30^\circ) = \sin 30^\circ = \frac{1}{2}$.

The Pythagorean identities

For every θ where the functions are defined:

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

$1 + \tan^2\theta = \sec^2\theta$

$1 + \cot^2\theta = \csc^2\theta$

The conjugate trick for sec θ + tan θ

Factorising $\sec^2\theta - \tan^2\theta = 1$ gives:

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

So if $\sec\theta + \tan\theta = k$, then $\sec\theta - \tan\theta = \frac{1}{k}$. Adding and subtracting:

$\sec\theta = \frac{k^2 + 1}{2k}, \qquad \tan\theta = \frac{k^2 - 1}{2k}, \qquad \sin\theta = \frac{k^2 - 1}{k^2 + 1}$

The same idea works for $\csc\theta \pm \cot\theta$. It is far quicker than squaring, and squaring can introduce extra roots.

Identities that come up repeatedly

The first two come from $a^2 + b^2 = (a + b)^2 - 2ab$ and $a^3 + b^3 = (a + b)^3 - 3ab(a + b)$ with $a = \sin^2\theta$, $b = \cos^2\theta$.

Worked examples

Example 1: sec θ + tan θ = p

If $\sec\theta + \tan\theta = p$ with $p > 1$, find $\sin\theta$ and the quadrant of θ.

Step 1: By the conjugate identity, $\sec\theta - \tan\theta = \frac{1}{p}$.

Step 2: Adding and subtracting:

$\sec\theta = \frac{p^2 + 1}{2p}, \qquad \tan\theta = \frac{p^2 - 1}{2p}$

Step 3: $\sin\theta = \frac{\tan\theta}{\sec\theta} = \frac{p^2 - 1}{p^2 + 1}$.

Step 4: For $p > 1$, both sec θ and tan θ are positive, which happens only in quadrant I.

Example 2: a sum of squares of cosines

Evaluate $S = \cos^2 5^\circ + \cos^2 10^\circ + \cdots + \cos^2 85^\circ + \cos^2 90^\circ$.

Step 1: There are 18 terms ($5^\circ$ to $90^\circ$ in steps of $5^\circ$).

Step 2: Pair complementary angles: $\cos^2 85^\circ = \sin^2 5^\circ$, so $\cos^2 5^\circ + \cos^2 85^\circ = 1$. The pairs $(5^\circ, 85^\circ), (10^\circ, 80^\circ), \ldots, (40^\circ, 50^\circ)$ give 8 pairs, each equal to 1.

Step 3: The unpaired terms are $\cos^2 45^\circ = \frac{1}{2}$ and $\cos^2 90^\circ = 0$.

$S = 8 + \frac{1}{2} + 0 = \frac{17}{2}$

Example 3: an expression that does not depend on x

Find $3(\sin x - \cos x)^4 + 6(\sin x + \cos x)^2 + 4(\sin^6 x + \cos^6 x)$.

Step 1: Let $t = \sin x\cos x$. Then $(\sin x \pm \cos x)^2 = 1 \pm 2t$.

Step 2: Expand each term:

$3(1 - 2t)^2 = 3 - 12t + 12t^2$

$6(1 + 2t) = 6 + 12t$

$4(1 - 3t^2) = 4 - 12t^2$

Step 3: Add: every $t$ term cancels, leaving $3 + 6 + 4 = 13$.

Check: At $x = 0$: $3(1) + 6(1) + 4(1) = 13$ ✓. When an answer is a constant, substituting a convenient angle (one that keeps every denominator non-zero) is a fast way to confirm it.

Summary table

Exam tips

• Remove full turns of $360^\circ$ first, then apply the allied-angle steps.
• Take the sign from the original function, not the co-function you switch to.
• Use the conjugate trick instead of squaring whenever sec ± tan or cosec ± cot is given.
• When the options are constants, confirm your answer by substituting $0^\circ$, $30^\circ$ or $45^\circ$ — but never an angle that makes a denominator zero.

For exam-style practice on these ideas, see NIMCET trigonometric ratios and identities PYQs and trigonometry shortcuts for NIMCET. All trigonometry articles are listed in the NIMCET Trigonometry chapter.

Common mistakes

• Taking the sign from the co-function: $\cos(90^\circ + \theta) = -\sin\theta$ because cos is negative in quadrant II.
• Writing $\sqrt{1 - \sin^2\theta} = \cos\theta$. In general it equals $|\cos\theta|$, which is $-\cos\theta$ in quadrants II and III.
• Keeping impossible values: in $2\sin^2\theta - 5\sin\theta + 2 = 0$, the root $\sin\theta = 2$ must be rejected; only $\sin\theta = \frac{1}{2}$ is valid.
• Using degrees in $l = r\theta$ or in calculus formulas.

Frequently Asked Questions

What does the ASTC rule mean?

It gives the positive ratios in each quadrant: All in the first, Sin (and cosec) in the second, Tan (and cot) in the third, and Cos (and sec) in the fourth.

When does a trigonometric function change to its co-function?

When the angle is an odd multiple of 90° plus or minus θ, such as 90° + θ or 270° − θ. For even multiples, such as 180° ± θ, the function stays the same.

Is sin 1 greater than sin 1°?

Yes. Without a degree sign, 1 means 1 radian, about 57.3°. Since sine increases from 0° to 90°, sin 1 is greater than sin 1°.

If sec θ + tan θ = k, what is sec θ − tan θ?

It is 1/k, because (sec θ + tan θ)(sec θ − tan θ) = sec²θ − tan²θ = 1.

What are the three Pythagorean identities?

sin²θ + cos²θ = 1, 1 + tan²θ = sec²θ and 1 + cot²θ = cosec²θ.