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Trigonometric Ratios, Identities & Allied Angles

Master Trigonometric Ratios Identities and Allied Angles for JEE Main JEE Advanced and Board Exams. Covers degree radian conversion, standard angle values, ASTC rule, Pythagorean identities, allied angle transformations, periodicity and graphs of trig functions, range of trig expressions with solved JEE questions.

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Trigonometry is the mathematics of angles and triangles — and it underpins vast areas of mathematics, physics, and engineering. At the heart of trigonometry are the six trigonometric functions — sine, cosine, tangent, and their reciprocals — and the rich network of identities that connect them. Mastering these identities is not just about memorisation; it is about developing fluency in a language that appears throughout calculus, coordinate geometry, complex numbers, and waves. For JEE (Main & Advanced), trigonometric identities and their applications form a consistently tested topic, appearing both directly (2–3 questions) and embedded in problems across other chapters.

1. Angle Measurement — Degrees and Radians

Angles can be measured in degrees or radians. The radian is the SI unit and is preferred in calculus.

$π radians = 180 °$ $\Rightarrow$ $1 radian = \frac{180 °}{π} \approx 57.3 °$

Conversion Formulae

Degrees to radians: multiply by $\frac{π}{180}$ Radians to degrees: multiply by $\frac{180}{π}$

Important Angle Conversions

Degrees	0°	30°	45°	60°	90°	120°	180°	270°	360°
Radians	$0$	$\frac{π}{6}$	$\frac{π}{4}$	$\frac{π}{3}$	$\frac{π}{2}$	$\frac{2 π}{3}$	$π$	$\frac{3 π}{2}$	$2 π$

Arc Length and Area of Sector

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

Arc length: $l = r θ$ Area of sector: $A = \frac{1}{2} r^{2} θ = \frac{1}{2} r l$

2. Trigonometric Ratios — Definitions and Signs

For a point $P (x, y)$ on a circle of radius $r$ making angle $θ$ with the positive $x$ -axis:

$\sin θ = \frac{y}{r}$ $\cos θ = \frac{x}{r}$ $\tan θ = \frac{y}{x}$ $cosec θ = \frac{r}{y}$ $\sec θ = \frac{r}{x}$ $\cot θ = \frac{x}{y}$

Signs in Each Quadrant — ASTC Rule

"All Students Take Calculus" (or "Add Sugar To Coffee")

Quadrant	Angles	Positive Functions	Negative Functions
I	$0 ° < θ < 90 °$	All (sin, cos, tan, all)	None
II	$90 ° < θ < 180 °$	Sin (and cosec)	cos, tan, sec, cot
III	$180 ° < θ < 270 °$	Tan (and cot)	sin, cos, sec, cosec
IV	$270 ° < θ < 360 °$	Cos (and sec)	sin, tan, cosec, cot

Values of Trigonometric Functions at Standard Angles

$θ$	$0 °$	$30 °$	$45 °$	$60 °$	$90 °$
$\sin θ$	$0$	$\frac{1}{2}$	$\frac{1}{\sqrt{2}}$	$\frac{\sqrt{3}}{2}$	$1$
$\cos θ$	$1$	$\frac{\sqrt{3}}{2}$	$\frac{1}{\sqrt{2}}$	$\frac{1}{2}$	$0$
$\tan θ$	$0$	$\frac{1}{\sqrt{3}}$	$1$	$\sqrt{3}$	undefined
$cosec θ$	undefined	$2$	$\sqrt{2}$	$\frac{2}{\sqrt{3}}$	$1$
$\sec θ$	$1$	$\frac{2}{\sqrt{3}}$	$\sqrt{2}$	$2$	undefined
$\cot θ$	undefined	$\sqrt{3}$	$1$	$\frac{1}{\sqrt{3}}$	$0$

Memory trick for sin values: $\sin 0 ° = \frac{\sqrt{0}}{2}$ , $\sin 30 ° = \frac{\sqrt{1}}{2}$ , $\sin 45 ° = \frac{\sqrt{2}}{2}$ , $\sin 60 ° = \frac{\sqrt{3}}{2}$ , $\sin 90 ° = \frac{\sqrt{4}}{2}$ . The numerators follow $\sqrt{0}, \sqrt{1}, \sqrt{2}, \sqrt{3}, \sqrt{4}$ .

3. Fundamental Trigonometric Identities

Reciprocal Identities

$\sin θ \cdot cosec θ = 1$ $\cos θ \cdot \sec θ = 1$ $\tan θ \cdot \cot θ = 1$

Quotient Identities

$\tan θ = \frac{\sin θ}{\cos θ}$ $\cot θ = \frac{\cos θ}{\sin θ}$

Pythagorean Identities

$\sin^{2} θ + \cos^{2} θ = 1$

$1 + \tan^{2} θ = \sec^{2} θ$ $\Rightarrow$ $\sec^{2} θ - \tan^{2} θ = 1$

$1 + \cot^{2} θ = {cosec}^{2} θ$ $\Rightarrow$ ${cosec}^{2} θ - \cot^{2} θ = 1$

Key factorisations from Pythagorean identities:

$\sec^{2} θ - \tan^{2} θ = (\sec θ + \tan θ) (\sec θ - \tan θ) = 1$
If $\sec θ + \tan θ = k$ , then $\sec θ - \tan θ = 1 / k$
$\sin^{4} θ + \cos^{4} θ = 1 - 2 \sin^{2} θ \cos^{2} θ = 1 - \frac{\sin^{2} 2 θ}{2}$

4. Allied Angle Identities

Allied angles are angles of the form $n \cdot 90 ° \pm θ$ where $n$ is an integer. Two simple rules govern all allied angle transformations:

Rule 1 (Function change): If $n$ is odd, the function changes (sin ↔ cos, tan ↔ cot, sec ↔ cosec). If $n$ is even, the function stays the same.
Rule 2 (Sign): The sign is determined by the sign of the original function in the quadrant where $n \cdot 90 ° \pm θ$ lies (treating $θ$ as acute).

Expression	$\sin$	$\cos$	$\tan$	Rule applied
$- θ$ (or $360 ° - θ$ )	$- \sin θ$	$\cos θ$	$- \tan θ$	Even function: cos; Odd: sin, tan
$90 ° - θ$	$\cos θ$	$\sin θ$	$\cot θ$	$n = 1$ odd: function changes
$90 ° + θ$	$\cos θ$	$- \sin θ$	$- \cot θ$	$n = 1$ odd: function changes; QII: cos+, sin–
$180 ° - θ$	$\sin θ$	$- \cos θ$	$- \tan θ$	$n = 2$ even: same function; QII: sin+, cos–
$180 ° + θ$	$- \sin θ$	$- \cos θ$	$\tan θ$	$n = 2$ even: same function; QIII: tan+, sin–, cos–
$270 ° - θ$	$- \cos θ$	$- \sin θ$	$\cot θ$	$n = 3$ odd: function changes; QIII: tan+
$270 ° + θ$	$- \cos θ$	$\sin θ$	$- \cot θ$	$n = 3$ odd: function changes; QIV: cos+
$360 ° - θ$	$- \sin θ$	$\cos θ$	$- \tan θ$	Same as $- θ$ ; QIV: cos+

Key Special Values

$\sin (180 ° - θ) = \sin θ$ → $\sin 150 ° = \sin 30 ° = \frac{1}{2}$
$\cos (180 ° - θ) = - \cos θ$ → $\cos 120 ° = - \cos 60 ° = - \frac{1}{2}$
$\tan (180 ° + θ) = \tan θ$ → $\tan$ has period $180 °$
$\sin (90 ° + θ) = \cos θ$ → $\sin 120 ° = \cos 30 ° = \frac{\sqrt{3}}{2}$

5. Periodicity and Graphs of Trigonometric Functions

Function	Domain	Range	Period	Even/Odd
$\sin x$	$R$	$[- 1, 1]$	$2 π$	Odd: $\sin (- x) = - \sin x$
$\cos x$	$R$	$[- 1, 1]$	$2 π$	Even: $\cos (- x) = \cos x$
$\tan x$	$R ∖ {\frac{π}{2} + n π}$	$R$	$π$	Odd: $\tan (- x) = - \tan x$
$\cot x$	$R ∖ {n π}$	$R$	$π$	Odd
$\sec x$	$R ∖ {\frac{π}{2} + n π}$	$(- \infty, - 1] \cup [1, \infty)$	$2 π$	Even
$cosec x$	$R ∖ {n π}$	$(- \infty, - 1] \cup [1, \infty)$	$2 π$	Odd

Period of Composite Functions

Period of $\sin (n θ)$ or $\cos (n θ)$ = $\frac{2 π}{n}$
Period of $\tan (n θ)$ or $\cot (n θ)$ = $\frac{π}{n}$
Period of $| \sin θ |$ or $| \cos θ |$ = $π$
Period of $| \tan θ |$ = $π$
Period of $\sin^{2} θ$ = $π$ (since $\sin^{2} θ = \frac{1 - \cos 2 θ}{2}$ )

6. Fundamental Identities — Proof Techniques

Approach for Proving Identities

Convert everything to sin and cos — the most reliable method. Express tan, cot, sec, cosec in terms of sin and cos, then simplify.
Work on the more complex side — simplify the LHS or RHS (whichever is more complex) to match the other.
Use Pythagorean identities — replace $1 - \sin^{2} θ$ with $\cos^{2} θ$ , etc.
Factorisation and algebraic manipulation — difference of squares, perfect squares, rationalisation.

Worked Example — Prove $\frac{\sec θ - 1}{\sec θ + 1} = \frac{1 - \cos θ}{1 + \cos θ}$

LHS $= \frac{\sec θ - 1}{\sec θ + 1}$
Multiply numerator and denominator by $\cos θ$ :
$= \frac{\cos θ (\sec θ - 1)}{\cos θ (\sec θ + 1)} = \frac{1 - \cos θ}{1 + \cos θ}$ = RHS ✓

Useful Algebraic Identities in Trigonometry

$\sin^{2} θ = \frac{1 - \cos 2 θ}{2}$ $\cos^{2} θ = \frac{1 + \cos 2 θ}{2}$
$(\sin θ + \cos θ)^{2} = 1 + 2 \sin θ \cos θ = 1 + \sin 2 θ$
$(\sin θ - \cos θ)^{2} = 1 - 2 \sin θ \cos θ = 1 - \sin 2 θ$
$\sin^{6} θ + \cos^{6} θ = 1 - 3 \sin^{2} θ \cos^{2} θ = 1 - \frac{3}{4} \sin^{2} 2 θ$

7. Range of Trigonometric Expressions

Finding the range of expressions like $a \sin θ + b \cos θ$ is a key JEE skill:

$- \sqrt{a^{2} + b^{2}} \leq a \sin θ + b \cos θ \leq \sqrt{a^{2} + b^{2}}$

This follows because $a \sin θ + b \cos θ = \sqrt{a^{2} + b^{2}} \sin (θ + ϕ)$ where $\tan ϕ = b / a$ .

Range of Other Common Expressions

Expression	Range
$a \sin θ + b \cos θ$	$[- \sqrt{a^{2} + b^{2}}, \sqrt{a^{2} + b^{2}}]$
$a \sin^{2} θ + b \cos^{2} θ$	$[min (a, b), max (a, b)]$
$\sin θ + \cos θ$	$[- \sqrt{2}, \sqrt{2}]$ ; max at $θ = 45 °$
$\sin θ \cdot \cos θ$	$[- \frac{1}{2}, \frac{1}{2}]$ (since $= \frac{\sin 2 θ}{2}$ )
$\sin^{2} θ + {cosec}^{2} θ$	$[2, \infty)$ (by AM-GM: $\geq 2$ )

Allied Angles — The Two-Step Rule (No Memorisation Needed)

For any expression of the form $f (n \cdot 90 ° \pm θ)$ :

Step	Rule	Example: $\sin (270 ° + θ)$
1. Function	$n$ odd → change function (sin↔cos, tan↔cot, sec↔cosec). $n$ even → keep same function.	$n = 3$ (odd) → sin becomes cos
2. Sign	Identify which quadrant $270 ° + θ$ lies in (treating $θ$ as small acute). Use ASTC for the sign of the resulting function.	$270 ° + θ$ is in QIV. In QIV, cos is positive. So $\sin (270 ° + θ) = - \cos θ$ . Wait — cos is positive in QIV but the resulting function is cos and we need its sign there: $+ \cos θ$ . Hmm — recheck: $\sin (270 ° + θ)$ . Quadrant IV — sin is negative there. Since $n$ is odd, sin→cos. But the sign comes from sin in QIV: sin is negative in QIV. So answer: $- \cos θ$ . ✓

Cleaner version of Step 2: After determining the new function from Step 1, the sign is determined by asking: "What is the sign of the original function in the quadrant where the angle $n \cdot 90 ° \pm θ$ lies?"

Example: $\sin (270 ° + θ)$ → $n = 3$ (odd) → function becomes cos → angle is in QIV → sin is negative in QIV → result is $- \cos θ$ .

JEE Power Tips — Avoid These Common Mistakes

$\tan$ and $\cot$ have period $π$ , not $2 π$ . A very common error — $\tan (180 ° + θ) = \tan θ$ , not $- \tan θ$ . Only sin, cos, sec, cosec have period $2 π$ .
$\cos$ is an even function; sin, tan are odd. So $\cos (- θ) = \cos θ$ (no sign change), but $\sin (- θ) = - \sin θ$ and $\tan (- θ) = - \tan θ$ .
Range of $a \sin θ + b \cos θ$ is $[- \sqrt{a^{2} + b^{2}}, \sqrt{a^{2} + b^{2}}]$ — not $[- (| a | + | b |), | a | + | b |]$ . The maximum of a sum is NOT the sum of the maxima.
$\sec^{2} θ - \tan^{2} θ = 1$ , not $\sec^{2} θ = 1 + \tan^{2} θ$ being misapplied. These are the same identity — but students often write $\sec^{2} θ - 1 = \tan^{2} θ$ and mistakenly use it as $\sec^{2} θ + 1 = \tan^{2} θ$ .
$\sin^{- 1} (\sin θ) \neq θ$ always. The inverse trig functions have restricted domains. $\sin^{- 1} (\sin 150 °) = \sin^{- 1} (1 / 2) = 30 °$ , not $150 °$ .
Allied angle rule — the sign uses the original function's sign, not the new function's sign. For $\cos (90 ° + θ)$ : $n = 1$ odd → cos becomes sin → angle is in QII → cos is negative in QII → result is $- \sin θ$ . The sign is from cos (original) in QII.
$| \sin θ |$ and $\sin^{2} θ$ both have period $π$ — not $2 π$ . This is a frequently tested MCQ on periodicity in JEE.

Practice Questions (JEE Level)

Q1: The value of $\sin 135^{\circ} + \cos 120^{\circ} + \cos 150^{\circ}$ is:

A) $\frac{\sqrt{2} - \sqrt{3} - 1}{2}$
B) $\frac{\sqrt{2} - \sqrt{3} + 1}{2}$
C) $\frac{1 - \sqrt{2} - \sqrt{3}}{2}$
D) $\frac{\sqrt{2} + \sqrt{3} - 1}{2}$

Answer: A) $\frac{\sqrt{2} - \sqrt{3} - 1}{2}$ .

Explanation:

Evaluate each trigonometric term using standard reference angles:
$\sin 135^{\circ} = \sin (180^{\circ} - 45^{\circ}) = \sin 45^{\circ} = \frac{\sqrt{2}}{2}$
$\cos 120^{\circ} = \cos (180^{\circ} - 60^{\circ}) = - \cos 60^{\circ} = - \frac{1}{2}$
$\cos 150^{\circ} = \cos (180^{\circ} - 30^{\circ}) = - \cos 30^{\circ} = - \frac{\sqrt{3}}{2}$

Add the values together:
$Sum = \frac{\sqrt{2}}{2} - \frac{1}{2} - \frac{\sqrt{3}}{2}$

Combine over the common denominator:
$Sum = \frac{\sqrt{2} - 1 - \sqrt{3}}{2}$

This can be rearranged to perfectly match Option A: $\frac{\sqrt{2} - \sqrt{3} - 1}{2}$ .

Q2: If $\sin θ + \cos θ = \sqrt{2} \cos θ$ , then $\cos θ - \sin θ$ equals:

A) $\sqrt{2} \sin θ$
B) $- \sqrt{2} \sin θ$
C) $\sqrt{2} \cos θ$
D) $\sqrt{2} \tan θ$

Answer: A) $\sqrt{2} \sin θ$ .

Explanation:

Square both sides of the given equation:
$(\sin θ + \cos θ)^{2} = (\sqrt{2} \cos θ)^{2}$
$\sin^{2} θ + \cos^{2} θ + 2 \sin θ \cos θ = 2 \cos^{2} θ$
$1 + 2 \sin θ \cos θ = 2 \cos^{2} θ$

Now, expand the target expression squared:
$(\cos θ - \sin θ)^{2} = \cos^{2} θ + \sin^{2} θ - 2 \sin θ \cos θ$
$(\cos θ - \sin θ)^{2} = 1 - 2 \sin θ \cos θ$

Substitute $2 \sin θ \cos θ = 2 \cos^{2} θ - 1$ from the first step:
$(\cos θ - \sin θ)^{2} = 1 - (2 \cos^{2} θ - 1)$
$(\cos θ - \sin θ)^{2} = 2 - 2 \cos^{2} θ = 2 (1 - \cos^{2} θ) = 2 \sin^{2} θ$

Taking the positive square root (since $\cos θ > \sin θ$ given the initial condition) gives:
$\cos θ - \sin θ = \sqrt{2} \sin θ$ .

Q3: The fundamental period of $f (x) = \sin^{2} x + \cos^{4} x$ is:

A) $2 π$
B) $π$
C) $π / 2$
D) $π / 4$

Answer: C) $π / 2$ .

Explanation:

First, simplify the function using trigonometric identities:
$f (x) = \sin^{2} x + \cos^{4} x$
$f (x) = 1 - \cos^{2} x + \cos^{4} x$
$f (x) = 1 - \cos^{2} x (1 - \cos^{2} x)$
$f (x) = 1 - \cos^{2} x \sin^{2} x$

Multiply and divide by 4 to use the double angle identity ( $\sin 2 x = 2 \sin x \cos x$ ):
$f (x) = 1 - \frac{1}{4} (4 \sin^{2} x \cos^{2} x)$
$f (x) = 1 - \frac{1}{4} (\sin 2 x)^{2} = 1 - \frac{1}{4} \sin^{2} (2 x)$

Use the power-reducing identity $\sin^{2} θ = \frac{1 - \cos 2 θ}{2}$ :
$f (x) = 1 - \frac{1}{4} (\frac{1 - \cos 4 x}{2})$
$f (x) = 1 - \frac{1}{8} + \frac{1}{8} \cos 4 x$
$f (x) = \frac{7}{8} + \frac{1}{8} \cos 4 x$

The period of $\cos (k x)$ is $\frac{2 π}{k}$ . Here, $k = 4$ , so the period is:
$Period = \frac{2 π}{4} = \frac{π}{2}$ .

Q4: If $\sec θ + \tan θ = k$ , then $\sin θ$ equals:

A) $\frac{k^{2} + 1}{k^{2} - 1}$
B) $\frac{k^{2} - 1}{k^{2} + 1}$
C) $\frac{2 k}{k^{2} + 1}$
D) $\frac{k}{k^{2} - 1}$

Answer: B) $\frac{k^{2} - 1}{k^{2} + 1}$ .

Explanation:

We know the standard identity: $\sec^{2} θ - \tan^{2} θ = 1$ .
This factors to: $(\sec θ + \tan θ) (\sec θ - \tan θ) = 1$ .

Substitute the given value ( $\sec θ + \tan θ = k$ ):
$k (\sec θ - \tan θ) = 1 ⟹ \sec θ - \tan θ = \frac{1}{k}$ .

Now, set up a system of equations:
1) $\sec θ + \tan θ = k$
2) $\sec θ - \tan θ = \frac{1}{k}$

Add equations 1 and 2 to find $\sec θ$ :
$2 \sec θ = k + \frac{1}{k} = \frac{k^{2} + 1}{k} ⟹ \sec θ = \frac{k^{2} + 1}{2 k}$

Subtract equation 2 from 1 to find $\tan θ$ :
$2 \tan θ = k - \frac{1}{k} = \frac{k^{2} - 1}{k} ⟹ \tan θ = \frac{k^{2} - 1}{2 k}$

Finally, find $\sin θ$ using the identity $\sin θ = \frac{\tan θ}{\sec θ}$ :
$\sin θ = \frac{\frac{k^{2} - 1}{2 k}}{\frac{k^{2} + 1}{2 k}} = \frac{k^{2} - 1}{k^{2} + 1}$ .

Q5: The maximum value of $3 \sin θ - 4 \cos θ$ is:

A) 3
B) 4
C) 5
D) 7

Answer: C) 5.

Explanation:

For any trigonometric expression in the form $a \sin θ + b \cos θ$ , the maximum possible value is given by the formula $\sqrt{a^{2} + b^{2}}$ .

In this expression, $a = 3$ and $b = - 4$ .

Calculate the maximum value:
$Max = \sqrt{3^{2} + (- 4)^{2}} = \sqrt{9 + 16} = \sqrt{25} = 5$ .

Q6: The exact value of $\csc (75^{\circ})$ is:

A) $\frac{2 + \sqrt{3}}{\sqrt{6} - \sqrt{2}}$
B) $\sqrt{6} - \sqrt{2}$
C) $\sqrt{6} + \sqrt{2}$
D) $\frac{4}{\sqrt{6} - \sqrt{2}}$

Answer: B) $\sqrt{6} - \sqrt{2}$ .

Explanation:

By definition, $\csc (75^{\circ}) = \frac{1}{\sin (75^{\circ})}$ .

First, find $\sin (75^{\circ})$ using the angle addition formula:
$\sin (45^{\circ} + 30^{\circ}) = \sin 45^{\circ} \cos 30^{\circ} + \cos 45^{\circ} \sin 30^{\circ}$
$= (\frac{\sqrt{2}}{2}) (\frac{\sqrt{3}}{2}) + (\frac{\sqrt{2}}{2}) (\frac{1}{2}) = \frac{\sqrt{6} + \sqrt{2}}{4}$

Therefore, $\csc (75^{\circ}) = \frac{4}{\sqrt{6} + \sqrt{2}}$ .

Rationalize the denominator by multiplying the numerator and denominator by the conjugate $(\sqrt{6} - \sqrt{2})$ :
$= \frac{4 (\sqrt{6} - \sqrt{2})}{(\sqrt{6} + \sqrt{2}) (\sqrt{6} - \sqrt{2})}$
$= \frac{4 (\sqrt{6} - \sqrt{2})}{6 - 2}$
$= \frac{4 (\sqrt{6} - \sqrt{2})}{4}$
$= \sqrt{6} - \sqrt{2}$ .

Q7 (JEE type): If $\sin θ + \csc θ = 2$ , then $\sin^{10} θ + \csc^{10} θ$ equals:

A) 2
B) $2^{10}$
C) 10
D) $2^{9}$

Answer: A) 2.

Explanation:

We are given $\sin θ + \csc θ = 2$ . Since $\csc θ = \frac{1}{\sin θ}$ , we can write:
$\sin θ + \frac{1}{\sin θ} = 2$

Multiply the entire equation by $\sin θ$ to form a quadratic equation:
$\sin^{2} θ + 1 = 2 \sin θ$
$\sin^{2} θ - 2 \sin θ + 1 = 0$

Factor the perfect square:
$(\sin θ - 1)^{2} = 0 ⟹ \sin θ = 1$ .

If $\sin θ = 1$ , then $\csc θ = \frac{1}{1} = 1$ .

Evaluate the target expression:
$\sin^{10} θ + \csc^{10} θ = (1)^{10} + (1)^{10} = 1 + 1 = 2$ .

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Trigonometric Ratios, Identities & Allied Angles

1. Angle Measurement — Degrees and Radians

Conversion Formulae

Important Angle Conversions

Arc Length and Area of Sector

2. Trigonometric Ratios — Definitions and Signs

Signs in Each Quadrant — ASTC Rule

Values of Trigonometric Functions at Standard Angles

3. Fundamental Trigonometric Identities

Reciprocal Identities

Quotient Identities

Pythagorean Identities

4. Allied Angle Identities

Key Special Values

5. Periodicity and Graphs of Trigonometric Functions

Period of Composite Functions

6. Fundamental Identities — Proof Techniques

Approach for Proving Identities

Worked Example — Prove $\frac{\sec θ - 1}{\sec θ + 1} = \frac{1 - \cos θ}{1 + \cos θ}$

Useful Algebraic Identities in Trigonometry

7. Range of Trigonometric Expressions

Range of Other Common Expressions

Allied Angles — The Two-Step Rule (No Memorisation Needed)

Practice Questions (JEE Level)

Related Study Material

Practice Questions

Topic Information

Quick Actions

Track Your Learning

Study Tips

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Target Subjects

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Subject Mastery

Trigonometric Ratios, Identities & Allied Angles

1. Angle Measurement — Degrees and Radians

Conversion Formulae

Important Angle Conversions

Arc Length and Area of Sector

2. Trigonometric Ratios — Definitions and Signs

Signs in Each Quadrant — ASTC Rule

Values of Trigonometric Functions at Standard Angles

3. Fundamental Trigonometric Identities

Reciprocal Identities

Quotient Identities

Pythagorean Identities

4. Allied Angle Identities

Key Special Values

5. Periodicity and Graphs of Trigonometric Functions

Period of Composite Functions

6. Fundamental Identities — Proof Techniques

Approach for Proving Identities

Worked Example — Prove sec⁡θ−1sec⁡θ+1=1−cos⁡θ1+cos⁡θ

Useful Algebraic Identities in Trigonometry

7. Range of Trigonometric Expressions

Range of Other Common Expressions

Allied Angles — The Two-Step Rule (No Memorisation Needed)

Practice Questions (JEE Level)

Related Study Material

Practice Questions

Topic Information

Quick Actions

Track Your Learning

Study Tips

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We Value Your Privacy

Worked Example — Prove $\frac{\sec θ - 1}{\sec θ + 1} = \frac{1 - \cos θ}{1 + \cos θ}$