Trigonometry & Geometry Study Notes

📐 Mathematics · Trigonometry & Geometry 🔥 Day 1

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Trigonometry & Inverse Trigonometric Functions

Identities · Compound angles · Equations · Inverse trig domains & identities

Standard Values & ASTC Rule

Angle	sin	cos	tan	cosec	sec	cot
0°	0	1	0	—	1	—
30° (π/6)	$\dfrac{1}{2}$	$\dfrac{\sqrt3}{2}$	$\dfrac{1}{\sqrt3}$	$2$	$\dfrac{2}{\sqrt3}$	$\sqrt3$
45° (π/4)	$\dfrac{1}{\sqrt2}$	$\dfrac{1}{\sqrt2}$	$1$	$\sqrt2$	$\sqrt2$	$1$
60° (π/3)	$\dfrac{\sqrt3}{2}$	$\dfrac{1}{2}$	$\sqrt3$	$\dfrac{2}{\sqrt3}$	$2$	$\dfrac{1}{\sqrt3}$
90° (π/2)	1	0	—	1	—	0

💡ASTC rule (quadrant signs): Q1 All positive; Q2 Sin positive; Q3 Tan positive; Q4 Cos positive. Mnemonic: "All Students Take Calculus." For angles beyond 90°: sin(180°−θ)=sinθ, cos(180°−θ)=−cosθ, tan(180°+θ)=tanθ, sin(−θ)=−sinθ, cos(−θ)=cosθ.

Fundamental Identities

Pythagorean Identities

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

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

$1 + \cot^2\theta = \csc^2\theta$ (→ $\csc^2\theta - \cot^2\theta = 1$)

Reciprocal & Quotient

$\sin\theta\cdot\csc\theta = 1$; $\cos\theta\cdot\sec\theta = 1$; $\tan\theta\cdot\cot\theta = 1$

$\tan\theta = \sin\theta/\cos\theta$; $\cot\theta = \cos\theta/\sin\theta$

Compound & Multiple Angle Formulas

Compound Angles

$\sin(A\pm B) = \sin A\cos B \pm \cos A\sin B$

$\cos(A\pm B) = \cos A\cos B \mp \sin A\sin B$

$\tan(A\pm B) = (\tan A \pm \tan B) / (1 \mp \tan A\tan B)$

Double Angle

$$ \sin 2A = 2\sin A\cos A = 2\tan A/(1+\tan^{2}A) $$

$$ \cos 2A = \cos^{2}A-\sin^{2}A = 2\cos^{2}A-1 = 1-2\sin^{2}A = \frac{1-\tan^{2}A}{1+\tan^{2}A} $$

$\tan 2A = 2\tan A/(1-\tan^2 A)$

Triple Angle

$\sin 3A = 3\sin A - 4\sin^3 A$

$\cos 3A = 4\cos^3 A - 3\cos A$

$\tan 3A = \dfrac{3\tan A - \tan^3 A}{1 - 3\tan^2 A}$

Half Angle (t-formulae), t = tan(A/2)

$$ sinA = \frac{2t}{1+t^{2}} $$

$$ cosA = \frac{1-t^{2}}{1+t^{2}} $$

$$ tanA = \frac{2t}{1-t^{2}} $$

Used to convert trig equations to algebraic (rational in t).

Sum-to-Product & Product-to-Sum

Sum → Product

sinC + sinD = 2 sin((C+D)/2) cos((C−D)/2)

sinC − sinD = 2 cos((C+D)/2) sin((C−D)/2)

cosC + cosD = 2 cos((C+D)/2) cos((C−D)/2)

cosC − cosD = −2 sin((C+D)/2) sin((C−D)/2)

Product → Sum

2 sinA cosB = sin(A+B) + sin(A−B)

2 cosA sinB = sin(A+B) − sin(A−B)

2 cosA cosB = cos(A−B) + cos(A+B)

2 sinA sinB = cos(A−B) − cos(A+B)

General Solutions of Trig Equations

Equation	General Solution
$\sin\theta = \sin\alpha$	$\theta = n\pi + (-1)^{n}\alpha, n \in ℤ$
$\cos\theta = \cos\alpha$	$\theta = 2n\pi \pm \alpha, n \in ℤ$
$\tan\theta = \tan\alpha$	$\theta = n\pi + \alpha, n \in ℤ$
$\sin\theta = 0$	$\theta = n\pi$
$\cos\theta = 0$	$\theta = (2n+1)\dfrac{\pi}{2}$
$\tan\theta = 0$	$\theta = n\pi$
$\sin^{2}\theta = \sin^{2}\alpha$	$\theta = n\pi \pm \alpha$ (same as cos²θ = cos²α and tan²θ = tan²α)

💡Extraneous solutions: When squaring both sides or using t = tan(A/2), always check solutions in the original equation. t-substitution excludes θ = (2n+1)π (where tanθ/2 is undefined).

Inverse Trigonometric Functions

Function	Domain	Range (principal)	Monotonicity
sin⁻¹ x	[−1, 1]	[−π/2, π/2]	Increasing
cos⁻¹ x	[−1, 1]	[0, π]	Decreasing
tan⁻¹ x	(−∞, ∞)	(−π/2, π/2)	Increasing
cot⁻¹ x	(−∞, ∞)	(0, π)	Decreasing
sec⁻¹ x	(−∞,−1]∪[1,∞)	[0,π]\{π/2}	Increasing on [1,∞)
cosec⁻¹ x	(−∞,−1]∪[1,∞)	[−π/2,π/2]\{0}	Decreasing on [1,∞)

Key Inverse Trig Identities

Complementary Relations

$\sin^{-1}x + \cos^{-1}x = \dfrac{\pi}{2}$ ($|x| \le 1$)

$\tan^{-1}x + \cot^{-1}x = \dfrac{\pi}{2}$ ($x \in \mathbb{R}$)

$\sec^{-1}x + \csc^{-1}x = \dfrac{\pi}{2}$ ($|x| \ge 1$)

Negative Argument

$\sin^{-1}(-x) = -\sin^{-1}x$

$\tan^{-1}(-x) = -\tan^{-1}x$

$$ \cos^{-1}(-x) = \pi - \cos^{-1}x $$

$$ \cot^{-1}(-x) = \pi - \cot^{-1}x $$

Addition Formulas

$\tan^{-1}x + \tan^{-1}y = \tan^{-1}(\dfrac{x+y}{1-xy})$ if $xy < 1$

$= \pi + \tan^{-1}(\dfrac{x+y}{1-xy})$ if $xy > 1$, $x > 0, y > 0$

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

Conversion Identities

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

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

$\sin^{-1}x + \sin^{-1}y = \sin^{-1}(x\sqrt{1-y^2} + y\sqrt{1-x^2})$ if $x^2+y^2\le1$

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Trigonometry & Inverse Trig — Key Formulas

Pythagorean Identities

$$ \sin^{2}\theta + \cos^{2}\theta = 1 \quad|\quad 1 + \tan^{2}\theta = \sec^{2}\theta \quad|\quad 1 + \cot^{2}\theta = \csc^{2}\theta $$

Compound Angles

sin(A±B) = sinAcosB ± cosAsinB
cos(A±B) = cosAcosB ∓ sinAsinB
tan(A±B) = (tanA±tanB)/(1∓tanAtanB)

Double Angle (three forms for cos 2A)

$\sin 2A = 2\sin A\cos A$
$\cos 2A = \cos^2 A - \sin^2 A = 2\cos^2 A - 1 = 1 - 2\sin^2 A$
$\tan 2A = 2\tan A/(1 - \tan^2 A)$

Triple Angle

$\sin 3A = 3\sin A - 4\sin^3 A$
$\cos 3A = 4\cos^3 A - 3\cos A$
$\tan 3A = \dfrac{3\tan A - \tan^3 A}{1 - 3\tan^2 A}$

Half Angle — t formulae (t = tan A/2)

$\sin A = \dfrac{2t}{1+t^2}$ | $\cos A = \dfrac{1-t^2}{1+t^2}$ | $\tan A = \dfrac{2t}{1-t^2}$

Sum-to-Product (most used pair)

sinC+sinD = 2sin((C+D)/2)cos((C−D)/2)
cosC+cosD = 2cos((C+D)/2)cos((C−D)/2)
cosC−cosD = −2sin((C+D)/2)sin((C−D)/2)

General Solutions

$\sin\theta=\sin\alpha \to \theta=n\pi+(-1)^n\alpha$
$\cos\theta=\cos\alpha \to \theta=2n\pi\pm\alpha$
$\tan\theta=\tan\alpha \to \theta=n\pi+\alpha$ ($n\in\mathbb Z$)

Inverse Trig — Complementary & Negative

sin⁻¹x+cos⁻¹x=π/2 | tan⁻¹x+cot⁻¹x=π/2
sin⁻¹(−x)=−sin⁻¹x | cos⁻¹(−x)=π−cos⁻¹x | tan⁻¹(−x)=−tan⁻¹x

tan⁻¹ Addition Formula

tan⁻¹x+tan⁻¹y = tan⁻¹((x+y)/(1−xy)) if xy<1
2tan⁻¹x = sin⁻¹(2x/(1+x²)) = cos⁻¹((1−x²)/(1+x²))

🧮

Trigonometry & Inverse Trig — Solved Examples

Find the exact value of sin 75° and cos 75°

Write 75° = 45° + 30° and apply compound angle formula:
sin 75° = sin(45°+30°) = sin45°cos30° + cos45°sin30°
$$ = (\frac{1}{\sqrt2})(\frac{\sqrt3}{2}) + (\frac{1}{\sqrt2})\frac{1}{2} = \frac{\sqrt3}{2\sqrt2} + \frac{1}{2\sqrt2} = $$ $$ \frac{\sqrt3+1}{2\sqrt2} = \frac{\sqrt6+\sqrt2}{4} $$
cos 75° = cos(45°+30°) = cos45°cos30° − sin45°sin30°
$= (\dfrac{1}{\sqrt2})(\dfrac{\sqrt3}{2}) - (\dfrac{1}{\sqrt2})\frac{1}{2} =$ $\dfrac{\sqrt6-\sqrt2}{4}$

Prove: $\dfrac{1 + \tan^{2}A}{1 + \cot^{2}A} = \tan^{2}A$

LHS = sec²A / cosec²A [using Pythagorean identities]
$= (\dfrac{1}{\cos^{2}}A) / (\dfrac{1}{\sin^{2}}A) = \dfrac{\sin^{2}A}{\cos^{2}}A =$ $\tan^{2}A = RHS$ ✓

Solve: $2\cos^2\theta + 3\sin\theta = 0$ for $\theta \in [0°, 360°]$

Replace cos²θ = 1−sin²θ:
$$ 2(1-\sin^{2}\theta) + 3\sin\theta = 0 \to 2\sin^{2}\theta - 3\sin\theta - 2 = 0 $$
Factoring: $(2\sin\theta + 1)(\sin\theta - 2) = 0$
sinθ = −1/2 (valid) or sinθ = 2 (impossible, |sinθ|≤1)
sinθ = −1/2: in [0°, 360°], θ = 210°, 330° (Q3 and Q4 where sin is negative)

Evaluate: tan⁻¹(1/2) + tan⁻¹(1/3)

Use $\tan^{-1}x + \tan^{-1}y = \tan^{-1}(\dfrac{x+y}{1-xy})$ when $xy < 1$.
$xy = (\dfrac{1}{2})(\dfrac{1}{3}) = \dfrac{1}{6} < 1$ ✓
$= \tan^{-1}(\dfrac{\dfrac{1}{2} + \dfrac{1}{3}}{1 - \dfrac{1}{6}}) = \tan^{-1}(\dfrac{\dfrac{5}{6}}{\dfrac{5}{6}}) = \tan^{-1}(1) =$ π/4

Simplify: sin⁻¹(sin 5π/6)

sin⁻¹ has principal range [−π/2, π/2]. We need to check if 5π/6 is in this range.
5π/6 ≈ 2.618 rad, which is outside [−π/2, π/2] = [−1.571, 1.571].
$$ \sin(5\frac{\pi}{6}) = \sin(\pi - 5\frac{\pi}{6}) = \sin(\frac{\pi}{6}) = \frac{1}{2} $$
$So \sin^{-1}(\sin 5\dfrac{\pi}{6}) = \sin^{-1}(\sin \pi/6) =$ π/6
Rule: sin⁻¹(sinθ) = θ only if θ ∈ [−π/2, π/2]; otherwise reduce first.

Find the range of $f(x) = \sin^{-1}x + \cos^{-1}x + \tan^{-1}x$

Domain: sin⁻¹x and cos⁻¹x require |x|≤1, while tan⁻¹x is defined for all x. So domain = [−1, 1].
Simplify: $\sin^{-1}x + \cos^{-1}x = \dfrac{\pi}{2}$ (identity for |x|≤1)
$$ So f(x) = \frac{\pi}{2} + \tan^{-1}x $$
Range of tan⁻¹x on [−1, 1]: tan⁻¹(−1) = −π/4; tan⁻¹(1) = π/4
Since tan⁻¹ is increasing: tan⁻¹x ∈ [−π/4, π/4] for x ∈ [−1, 1]
Therefore f(x) ∈ [π/2 − π/4, π/2 + π/4] = [π/4, 3π/4]

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Trigonometry & Inverse Trig — Key Diagrams

Click any diagram to zoom

Unit Circle — Standard Angles & ASTC Quadrant Signs

Unit circle showing key angles and quadrant sign rules. The ASTC mnemonic tells which functions are positive in each quadrant. Allied angle rule: for odd multiples of 90° (90°, 270°), sin↔cos flip; for even multiples (180°, 360°), function name stays, sign determined by ASTC.

Inverse Trig Functions — Domain & Range Visualization

sin⁻¹ is the restricted inverse of sin on [−π/2, π/2]. cos⁻¹ uses [0,π] so it is single-valued but decreasing. tan⁻¹ is defined everywhere but has horizontal asymptotes at ±π/2 — the curve never actually reaches those values.

General Solution — sinθ = sinα on the Number Line

The general solution $\theta = n\pi+(-1)^n\alpha$ captures all solutions of $\sin\theta=\sin\alpha$ on the number line. For even $n$ the formula gives $\alpha + 2k\pi$; for odd $n$ it gives $(\pi-\alpha) + 2k\pi$ — reflecting sin's symmetry about $\dfrac{\pi}{2}$.

⚡

Trigonometry & Inverse Trig — Quick Quiz

Question 1

The value of sin 15° equals:

$\sin15° = \sin(45°-30°) = \sin45°\cos30° - \cos45°\sin30° = (\dfrac{1}{\sqrt2})(\dfrac{\sqrt3}{2}) - (\dfrac{1}{\sqrt2})(\dfrac{1}{2}) = \dfrac{\sqrt3-1}{2\sqrt2}$. This equals $\dfrac{\sqrt6-\sqrt2}{4}$ after rationalising — option C is correct. Note: option D is sin75°.

Question 2

The general solution of $\tan\theta = -1$ is:

tanθ = tanα → θ = nπ+α. Here tanθ = −1 = tan(−π/4), so α = −π/4. General solution: θ = nπ+(−π/4) = nπ−π/4. Options A (nπ+π/4) gives tanθ=+1. Options C and D are for cosθ-type equations.

Question 3

The value of tan⁻¹(1) + tan⁻¹(2) + tan⁻¹(3) equals:

$\tan^{-1}2 + \tan^{-1}3$: $xy = 6 > 1$ with both positive, so $= \pi + \tan^{-1}(\dfrac{2+3}{1-6}) = \pi + \tan^{-1}(-1) = \pi - \dfrac{\pi}{4} = 3\dfrac{\pi}{4}$. Then $\tan^{-1}1 + 3\dfrac{\pi}{4} = \dfrac{\pi}{4} + 3\dfrac{\pi}{4} =$ π.

Question 4

Which of the following is TRUE?

A is false: $\cos^{-1}(-x) = \pi - \cos^{-1}(x)$, not the negative. B is false: $3\dfrac{\pi}{4}$ is outside $[-\dfrac{\pi}{2},\dfrac{\pi}{2}]$ — $\sin^{-1}(\sin 3\dfrac{\pi}{4}) = \sin^{-1}(\sin\pi/4) = \dfrac{\pi}{4}$. C is false: $2\dfrac{\pi}{3}$ is outside $(-\dfrac{\pi}{2},\dfrac{\pi}{2})$ — $\tan^{-1}(\tan 2\dfrac{\pi}{3}) = \tan^{-1}(\tan(2\dfrac{\pi}{3} - \pi)) = \tan^{-1}(\tan(-\dfrac{\pi}{3})) = -\dfrac{\pi}{3}$. D is the standard complementary identity — always true for $|x|\le 1$. ✓

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Straight Lines

Slopes · Line forms · Distance · Angle between lines · Pair of lines

Slope and Basic Concepts

Slope: $m = \tan\theta = \dfrac{y_2 - y_1}{x_2 - x_1}$. Horizontal line: $m = 0$. Vertical line: slope undefined.
Parallel lines: $m_{1} = m_{2}.$ Perpendicular lines: $m_{1}\cdot m_{2} = -1.$
Collinearity of 3 points: Area of triangle = 0, or determinant |x₁ y₁ 1; x₂ y₂ 1; x₃ y₃ 1| = 0.

Equations of a Line

Form	Equation	Use when
Slope-intercept	y = mx + c	Slope m and y-intercept c known
Point-slope	$y - y_{1} = m(x - x_{1})$	One point and slope known
Two-point	$\dfrac{y-y_{1}}{y_{2}-y_{1}} = \dfrac{x-x_{1}}{x_{2}-x_{1}}$	Two points known
Intercept form	x/a + y/b = 1	x-intercept a and y-intercept b known
Normal form	$x\cdot \cos\alpha + y\cdot \sin\alpha = p$	$p$ = ⊥ distance from origin; $\alpha$ = angle of normal
General form	ax + by + c = 0	Slope = −a/b; intercepts: −c/a, −c/b

Distances and Angles

Distance from point (x₁,y₁) to line ax+by+c=0: $d = |ax_{1}+by_{1}+c|/\sqrt{a^{2}+b^{2}}.$
Distance between parallel lines ax+by+c₁=0 and ax+by+c₂=0: $|c_1-c_2|/\sqrt{a^2+b^2}$.
Angle between two lines (slopes $m_1, m_2$): $\tan\theta = |m_1 - m_2|/(1 + m_1 m_2)$. $\theta = 90°$ if $m_1 m_2 = -1$.
Foot of perpendicular from $(x_1, y_1)$ to $ax + by + c = 0$:
$(x_1 - a\cdot t, y_1 - b\cdot t)$ where $t = \dfrac{ax_1 + by_1 + c}{a^2 + b^2}$.
Reflection of $(x_1, y_1)$ in line $ax + by + c = 0$: $(x_1 - 2at, y_1 - 2bt)$ with same $t$.
Area of triangle with vertices (x₁,y₁), (x₂,y₂), (x₃,y₃): ½|det of 3×2 matrix + col of 1s|.

Pair of Lines and Family of Lines

Pair of lines through origin: ax²+2hxy+by²=0. Two lines if h²≥ab; real and distinct if h²>ab; coincident if h²=ab.
For ax²+2hxy+by²=0: sum of slopes = −2h/b; product = a/b; angle: $\tan\theta = 2\dfrac{\sqrt{h^2-ab}}{a+b}$.
Lines perpendicular iff $a+b=0$ (coefficient of $x^2$ + coefficient of $y^2$ = 0).
Family of lines through intersection of $L_1 \equiv 0$ and $L_2 \equiv 0$: $L_1 + \lambda L_2 = 0$ for any $\lambda$. Pass through the common point for all $\lambda$.
Angle bisectors of ax+by+c=0 and a′x+b′y+c′=0: $\dfrac{ax+by+c}{\sqrt{a^2+b^2}} = \dfrac{\pm(a'x+b'y+c')}{\sqrt{a'^2+b'^2}}$.

💡 JEE Tip: For "foot of perpendicular" and "reflection", memorise the parameter $t = \dfrac{ax_1 + by_1 + c}{a^2 + b^2}$. Foot $= (x_1 - at, y_1 - bt)$; Image $= (x_1 - 2at, y_1 - 2bt)$. This one formula handles both. Also: angle bisector — use positive sign for bisector of acute angle, negative for obtuse angle bisector.

⚠ Common Error: Angle between lines: formula $\tan\theta = |\dfrac{m_1 - m_2}{1 + m_1 m_2}|$ gives the ACUTE angle. For the obtuse angle, use $\pi - \theta$. Also: when two lines are perpendicular, $m_1 m_2 = -1$ applies, but for vertical line (undefined slope), perpendicular is horizontal ($m = 0$) — don't apply the formula blindly.

🔢

Straight Lines — Formula Sheet

Distance Formulas

$$ d = |ax_{1}+by_{1}+c| / \sqrt{a^{2}+b^{2}} $$

Distance from point (x₁,y₁) to line ax+by+c=0. Parallel lines distance: $|c_1-c_2|/\sqrt{a^2+b^2}$.

$\tan\theta = |\dfrac{m_1 - m_2}{1 + m_1 m_2}|$ | $m_1 m_2 = -1$ (perp) | $m_1 = m_2$ (parallel)

Angle between two non-vertical lines. Parallel: same slope, different intercept. Perpendicular: product of slopes = −1.

Pair of lines: ax²+2hxy+by²=0 | $\tan\theta = 2\dfrac{\sqrt{h^2-ab}}{a+b}$

Perp iff $a+b=0$. Coincident iff $h^2=ab$. Sum of slopes $= -\dfrac{2h}{b}$; product $= \dfrac{a}{b}$.

🧮

Straight Lines — Solved Examples

Find: (a) distance from (3,−4) to 3x−4y+5=0; (b) find its foot of perpendicular on the line.

$d = |3(3)-4(-4)+5|/\sqrt{9+16} = |9+16+5|/5 = \dfrac{30}{5} =$ 6.
$t = \dfrac{ax_1 + by_1 + c}{a^2 + b^2} = \dfrac{9 + 16 + 5}{25} = \dfrac{30}{25} = \dfrac{6}{5}$. Foot $= (3 - 3\cdot(\dfrac{6}{5}),\, -4 - (-4)\cdot(\dfrac{6}{5})) = (3 - \dfrac{18}{5},\, -4 + \dfrac{24}{5}) = (-\dfrac{3}{5},\, \dfrac{4}{5})$.

Find the line through the intersection of 2x+y−3=0 and x−2y+1=0 that passes through (2,3).

Family: $(2x+y-3) + \lambda(x-2y+1) = 0 \to (2+\lambda)x + (1-2\lambda)y + (-3+\lambda) = 0.$
Passes through (2,3): $(2+\lambda)(2) + (1-2\lambda)(3) + (-3+\lambda) = 0 \to 4+2\lambda+3-6\lambda-3+\lambda = 0 \to 4-3\lambda=0 \to \lambda=\dfrac{4}{3.}$
Line: (2+4/3)x + (1−8/3)y + (−3+4/3) = 0 → (10/3)x + (−5/3)y + (−5/3) = 0 → 10x−5y−5=0 → 2x−y−1=0.

Find the image (reflection) of the point (1, 2) in the line $3x - 4y - 1 = 0$.

For the image $(x', y')$ of $(x_1, y_1)$ in $ax+by+c=0$: $\dfrac{x'-x_1}{a} = \dfrac{y'-y_1}{b} = \dfrac{-2(ax_1+by_1+c)}{a^2+b^2}.$
Here $a=3,\ b=-4,\ c=-1$; $ax_1+by_1+c = 3(1)-4(2)-1 = -6$. So the common ratio $= \dfrac{-2(-6)}{9+16} = \dfrac{12}{25}.$
$x' = 1 + 3\cdot\tfrac{12}{25} = 1 + \tfrac{36}{25} = \tfrac{61}{25};\quad y' = 2 + (-4)\cdot\tfrac{12}{25} = 2 - \tfrac{48}{25} = \tfrac{2}{25}.$

✓ Image $= \left(\dfrac{61}{25},\ \dfrac{2}{25}\right)$.

Find the equation of the bisector of the acute angle between the lines $3x - 4y + 7 = 0$ and $12x + 5y - 2 = 0$.

Bisectors: $\dfrac{3x-4y+7}{\sqrt{9+16}} = \pm\,\dfrac{12x+5y-2}{\sqrt{144+25}} \Rightarrow \dfrac{3x-4y+7}{5} = \pm\,\dfrac{12x+5y-2}{13}.$
Write both constant terms positive (they already are: $+7$ and $-2\to$ multiply 2nd line by $-1$ so $c>0$). Compute $a_1a_2+b_1b_2 = (3)(-12)+(-4)(-5) = -36+20 = -16 < 0$, so the positive sign gives the acute-angle bisector.
$13(3x-4y+7) = 5(-12x-5y+2) \Rightarrow 39x-52y+91 = -60x-25y+10 \Rightarrow 99x - 27y + 81 = 0.$

✓ Acute-angle bisector: $11x - 3y + 9 = 0$ (dividing by 9).

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Straight Lines — Visual Concepts

Click any diagram to zoom

Line Forms and Distance from Point to Line

The perpendicular distance from point P(x₁,y₁) to line ax+by+c=0 is $|ax_1+by_1+c|/\sqrt{a^2+b^2}$. The foot of the perpendicular lies at (x₁−at, y₁−bt) where t=(ax₁+by₁+c)/(a²+b²). The direction (a,b) is the normal to the line.

⚡

Straight Lines — Quick Quiz

Question 1

The distance from (0,0) to the line 3x+4y+10=0 is:

$d = |3(0)+4(0)+10|/\sqrt{9+16} = \dfrac{10}{\sqrt{25}} = \dfrac{10}{5} = 2$. Answer: C ✓

Question 2

Two lines are perpendicular. One has slope 3. The slope of the other is:

Perpendicular lines: m₁·m₂ = −1. So m₂ = −1/m₁ = −1/3. Answer: B ✓. Note: −3 would be a line parallel to the perpendicular direction; 1/3 has the same sign as the negative reciprocal would require.

Question 3

For the pair of lines ax²+2hxy+by²=0 to be perpendicular, the condition is:

For $ax^2 + 2hxy + by^2 = 0$: lines are perpendicular iff (sum of slopes)$^2 - 4$(product of slopes) $= -4$ is satisfied — but simpler: perpendicular iff $m_1 m_2 = -1$ → product $= \dfrac{a}{b} = -1$ → $a+b=0$. Answer: A ✓. Note: $h^2 = ab$ means lines coincide (not perpendicular).

⭕

Circles & Conic Sections

Circle · Parabola · Ellipse · Hyperbola · Tangents · Normals

Circle

Standard form: (x−h)²+(y−k)²=r², centre (h,k), radius r.
General form: x²+y²+2gx+2fy+c=0. Centre = (−g,−f); $r = \sqrt{g^2+f^2-c}$. Circle is real iff g²+f²−c > 0.
Tangent at (x₁,y₁) on circle (T=0): $xx_{1}+yy_{1}+g(x+x_{1})+f(y+y_{1})+c=0.$
Length of tangent from external point (x₁,y₁) to circle S=0: $\sqrt{S_1}$ where S₁ = x₁²+y₁²+2gx₁+2fy₁+c.
Two circles: radical axis = locus of points with equal tangent lengths to both circles (obtained by subtracting the two circle equations).

Parabola

Property	$y^{2}=4ax$ (opens right)	$x^{2}=4ay$ (opens up)
Vertex	(0,0)	(0,0)
Focus	(a,0)	(0,a)
Directrix	x = −a	y = −a
Latus rectum	x=a; length=4a	y=a; length=4a
Parametric	(at², 2at)	(2at, at²)
Tangent at (x₁,y₁)	$yy_{1} = 2a(x+x_{1})$	$xx_{1} = 2a(y+y_{1})$
Tangent (slope m)	y = mx + a/m	$y = mx - am^{2}$

Ellipse x²/a² + y²/b² = 1 (a > b > 0)

c² = a²−b², eccentricity $e = \dfrac{c}{a} = \sqrt{1-\dfrac{b^2}{a^2}}$, 0 < e < 1. Foci: (±ae, 0) = (±c, 0).
Sum of focal distances = 2a (defining property: PS₁ + PS₂ = 2a).
Directrices: $x = \pm \dfrac{a}{e}$. Latus rectum length $= \dfrac{2b^2}{a}$.
Tangent at (x₁,y₁): xx₁/a² + yy₁/b² = 1. Tangent (slope m): $y = mx \pm \sqrt{a^2m^2+b^2}$.
Parametric: $(a\cos\theta, b\sin\theta)$. At $\theta = 0$: $(a, 0)$ = right vertex.

Hyperbola x²/a² − y²/b² = 1

c² = a²+b², $e = \dfrac{c}{a} = \sqrt{1+\dfrac{b^2}{a^2}}$, e > 1. Foci: (±ae, 0).
|PS₁ − PS₂| = 2a (difference of focal distances).
Asymptotes: $y = \pm(\dfrac{b}{a})x$, or $bx - ay = 0$ and $bx + ay = 0$. Combined equation: $\dfrac{x^2}{a^2} - \dfrac{y^2}{b^2} = 0$.
Rectangular hyperbola: a=b → asymptotes at 90° → xy=c² form; $e=\sqrt2$.
Tangent at (x₁,y₁): xx₁/a²−yy₁/b²=1. Tangent (slope m): $y=mx\pm\sqrt{a^2m^2-b^2}$ (valid for |m|<b/a doesn't apply — valid when a²m²≠b²).

💡 JEE Tip: For any conic, "T=0" gives the equation of the tangent at point (x₁,y₁). To form T from the conic equation: replace x² with xx₁, y² with yy₁, 2x with (x+x₁), 2y with (y+y₁), xy with (xy₁+yx₁)/2. This rule works for circles, parabola, ellipse, and hyperbola.

⚠ Common Error: For ellipse x²/a²+y²/b²=1 with a>b: major axis is along x-axis (a is larger), foci are on x-axis at (±c,0). If b>a, foci are on y-axis. Don't confuse a and b — always identify which is larger to determine orientation. Eccentricity formula: for ellipse e=c/a<1; for hyperbola e=c/a>1.

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Circles & Conics — Formula Sheet

Conic Standard Forms Summary

Circle: (x−h)²+(y−k)²=r² | General: x²+y²+2gx+2fy+c=0, c(−g,−f), $r=\sqrt{g^2+f^2-c}$

Tangent at (x₁,y₁): T = xx₁+yy₁+g(x+x₁)+f(y+y₁)+c = 0. Length of tangent: $\sqrt{S_1}$.

Parabola y²=4ax: focus (a,0), directrix x=−a, T: yy₁=2a(x+x₁)

Slope form tangent: $y = mx + \dfrac{a}{m}$. Parametric: $(at^2, 2at)$. Latus rectum $= 4a$.

Ellipse: a²=b²+c²; e=c/a; foci (±c,0); LR=2b²/a; sum of focal dist=2a

Tangent (slope m): $y=mx\pm\sqrt{a^2m^2+b^2}$. At (x₁,y₁): xx₁/a²+yy₁/b²=1.

Hyperbola: c²=a²+b²; e=c/a>1; asymptotes y=±(b/a)x; |PS₁−PS₂|=2a

Rectangular hyperbola: xy=c², $e=\sqrt2$. Asymptotes perpendicular.

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Circles & Conics — Solved Examples

For the circle x²+y²−4x+6y−3=0: find centre, radius, and length of tangent from (5,0).

General form: 2g=−4→g=−2; 2f=6→f=3; c=−3. Centre = (−g,−f) = (2,−3).
$r = \sqrt{g^{2}+f^{2}-c} = \sqrt{4+9+3} = \sqrt{16} =$ 4.
Length of tangent from (5,0): $\sqrt{S_{1}} = \sqrt{25+0-20+0-3} = \sqrt2 =$ $\sqrt2$.

For the ellipse 9x²+16y²=144: find a, b, c, eccentricity, foci, and length of latus rectum.

Divide by 144: $\dfrac{x^2}{16} + \dfrac{y^2}{9} = 1$. $a^2 = 16 \to a = 4$; $b^2 = 9 \to b = 3$ ($a > b$, major axis along $x$).
$c^2 = a^2 - b^2 = 16 - 9 = 7$ → $c = \sqrt7$. $e = \dfrac{c}{a} = \dfrac{\sqrt7}{4}$.
Foci: $(\pm\sqrt7, 0)$. Latus rectum $= \dfrac{2b^2}{a} = 2\times \dfrac{9}{4} =$ 9/2.

Find the equation of tangent to y²=8x that is (a) parallel to y=2x+3; (b) perpendicular to y=2x+3.

y²=8x → 4a=8 → a=2. Tangent with slope m: y = mx + a/m = mx + 2/m.
(a) Parallel to y=2x+3 means m=2. Tangent: y = 2x + 2/2 = y = 2x+1.
(b) Perpendicular to y=2x+3 means slope m = −1/2. Tangent: y = −x/2 + 2/(−1/2) = y = −x/2 − 4, or x+2y+8=0.

✓ For parabola y²=4ax, slope form tangent y=mx+a/m is valid for all m≠0. The condition "c=a/m" ensures the line is tangent (touches at exactly one point).

For the hyperbola $9x^2 - 16y^2 = 144$: find $a,\ b$, eccentricity, foci, and the equations of the asymptotes.

Divide by 144: $\dfrac{x^2}{16} - \dfrac{y^2}{9} = 1$. So $a^2 = 16 \to a = 4$; $b^2 = 9 \to b = 3$.
For a hyperbola $b^2 = a^2(e^2 - 1)$: $9 = 16(e^2-1) \Rightarrow e^2 = 1 + \tfrac{9}{16} = \tfrac{25}{16} \Rightarrow e = \tfrac54$. Foci $(\pm ae, 0) = (\pm 5, 0)$.
Asymptotes: $\dfrac{x^2}{16} - \dfrac{y^2}{9} = 0 \Rightarrow y = \pm \tfrac{b}{a}x = \pm \tfrac34 x$, i.e. $3x \pm 4y = 0$.

✓ $a=4,\ b=3,\ e=\tfrac54$; foci $(\pm5,0)$; asymptotes $3x \pm 4y = 0$.

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Circles & Conics — Visual Concepts

Click any diagram to zoom

Conic Sections — Four Types with Eccentricity

Four conic sections classified by eccentricity. Circle (e=0): all points equidistant from centre. Ellipse (0<e<1): sum of focal distances constant. Parabola (e=1): equidistant from focus and directrix. Hyperbola (e>1): difference of focal distances constant; has two branches and two asymptotes.

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Circles & Conics — Quick Quiz

Question 1

For the circle x²+y²+4x−6y+4=0, the centre is:

2g=4→g=2; 2f=−6→f=−3. Centre = (−g,−f) = (−2, 3). Radius = $\sqrt{g^2+f^2-c} = \sqrt{4+9-4} = 3$. Answer: B ✓

Question 2

The eccentricity of a rectangular hyperbola is:

Rectangular hyperbola: a=b (asymptotes are perpendicular, at 90°). c²=a²+b²=2a². So $e=\dfrac{c}{a}=\sqrt2$. The equation xy=c² is the standard form of rectangular hyperbola (rotated). Answer: C ✓

Question 3

The tangent to the parabola y²=12x with slope 2 is:

y²=12x → 4a=12 → a=3. Tangent with slope m: y=mx+a/m = 2x+3/2. Answer: A ✓. Verify: substitute y=2x+3/2 into y²=12x: (2x+3/2)²=12x → 4x²+6x+9/4=12x → 4x²−6x+9/4=0 → discriminant=36−36=0 ✓ (tangent).

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Vectors & 3D Geometry

Dot & Cross Products · Triple Products · Lines & Planes in 3D · Skew Lines

Vector Algebra — Products

Dot product: a⃗·b⃗ = |a||b|cosθ = a₁b₁+a₂b₂+a₃b₃. Scalar result. a⃗·b⃗=0 iff perpendicular (a,b≠0).
Cross product: a⃗×b⃗ = |a||b|sinθ n̂. Vector result, perpendicular to both. |a⃗×b⃗|=0 iff parallel.
Cross product component form: a⃗×b⃗ = |i j k; a₁ a₂ a₃; b₁ b₂ b₃| (determinant form).
Areas: Area of parallelogram = |a⃗×b⃗|. Area of triangle = ½|a⃗×b⃗|.
Scalar triple product: [a⃗ b⃗ c⃗] = a⃗·(b⃗×c⃗) = volume of parallelepiped. [a⃗ b⃗ c⃗]=0 iff a⃗,b⃗,c⃗ are coplanar.
Properties: [a⃗ b⃗ c⃗] = [b⃗ c⃗ a⃗] = [c⃗ a⃗ b⃗] (cyclic); [a⃗ a⃗ b⃗] = 0; a⃗×(b⃗×c⃗) = (a⃗·c⃗)b⃗−(a⃗·b⃗)c⃗ (BAC−CAB).

3D Geometry — Lines

Direction cosines: $l=\cos\alpha, m=\cos\beta, n=\cos\gamma$ where α,β,γ are angles with x,y,z axes. l²+m²+n²=1.
Direction ratios (d.r.): a,b,c proportional to l,m,n. DC: $l=\dfrac{a}{\sqrt{a^2+b^2+c^2}}$, etc.
Line through point (x₁,y₁,z₁) with direction (a,b,c):
Cartesian: (x−x₁)/a = (y−y₁)/b = (z−z₁)/c = t.
Vector: r⃗ = a⃗₁ + t·b⃗.
Angle between two lines: $\cos\theta = |a_{1}a_{2}+b_{1}b_{2}+c_{1}c_{2}| / (\sqrt{a_{1}^{2}+b_{1}^{2}+c_{1}^{2}}\cdot\sqrt{a_{2}^{2}+b_{2}^{2}+c_{2}^{2}}).$
Skew lines: Non-intersecting, non-parallel lines in 3D. Shortest distance (SD) = |( b⃗₁−a⃗₁)·(d⃗₁×d⃗₂)| / |d⃗₁×d⃗₂|.

3D Geometry — Planes

Equation: ax+by+cz=d, where (a,b,c) is normal to the plane. Vector form: r⃗·n⃗=d.
Intercept form: x/p + y/q + z/r = 1 (intercepts p, q, r on axes).
Plane through 3 points: use the normal n⃗ = (P₂−P₁)×(P₃−P₁).
Distance from point (x₁,y₁,z₁) to plane ax+by+cz=d: $|ax_1+by_1+cz_1-d|/\sqrt{a^2+b^2+c^2}$.
Angle between two planes: $\cos\theta = |\vec{n}_{1}\cdot\vec{n}_{2}|/(|\vec{n}_{1}||\vec{n}_{2}|).$
Line-plane angle: sinφ = |a⃗·n⃗|/(|a⃗||n⃗|) where a⃗ is line direction, n⃗ is plane normal.

💡 JEE Tip: For shortest distance between skew lines, the formula SD = |(a⃗₂−a⃗₁)·(b⃗₁×b⃗₂)|/|b⃗₁×b⃗₂| requires the cross product of the two direction vectors. If b⃗₁×b⃗₂=0⃗, lines are parallel — use a different formula. Also: [a⃗ b⃗ c⃗]=0 is the key condition for coplanarity of three vectors.

⚠ Common Error: Direction cosines satisfy l²+m²+n²=1, but direction ratios (a,b,c) are just proportional — they need NOT satisfy a²+b²+c²=1. To convert: divide each by $\sqrt{a^2+b^2+c^2}$. Also: angle between a line and a plane uses SINE (sinφ = |d⃗·n̂|), not cosine — because the line's direction and the plane's normal are complementary angles.

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Vectors & 3D — Formula Sheet

Vector Products

a⃗·b⃗ = |a||b|cosθ = a₁b₁+a₂b₂+a₃b₃ | |a⃗×b⃗| = |a||b|sinθ

Dot: scalar, zero iff perpendicular. Cross: vector, zero iff parallel. Cross direction: right-hand rule.

[a b c] = a⃗·(b⃗×c⃗) = 0 iff coplanar | Volume of parallelepiped = |[a b c]|

Scalar triple product. Cyclic: [a b c]=[b c a]=[c a b]. [a a b]=0.

Lines and Planes

Skew line SD = |(a⃗₂−a⃗₁)·(b⃗₁×b⃗₂)| / |b⃗₁×b⃗₂|

a⃗₁, a⃗₂ = position vectors of points on lines. b⃗₁, b⃗₂ = direction vectors. SD=0 if intersecting.

Point-to-plane: $d = |ax_{1}+by_{1}+cz_{1}-d| / \sqrt{a^{2}+b^{2}+c^{2}}$

Same structure as 2D distance from point to line. Normal vector (a,b,c). Between parallel planes: $|d_1-d_2|/\sqrt{a^2+b^2+c^2}$.

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Vectors & 3D — Solved Examples

Find the area of triangle with vertices A(1,2,3), B(2,−1,4), C(0,1,2).

AB⃗ = B−A = (1,−3,1). AC⃗ = C−A = (−1,−1,−1).
AB⃗×AC⃗ = |i j k; 1 −3 1; −1 −1 −1| = i[(−3)(−1)−(1)(−1)] − j[(1)(−1)−(1)(−1)] + k[(1)(−1)−(−3)(−1)] = i[3+1]−j[−1+1]+k[−1−3] = (4,0,−4).
$|AB⃗×AC⃗| = \sqrt{16+0+16} = \sqrt{32} = 4\sqrt2$. Area = $½×4\sqrt2$ = $2\sqrt2$.

Find the shortest distance between lines: $L_{1}: \dfrac{x-1}{2}=\dfrac{y+1}{3}=\dfrac{z-1}{4}$ and $L_{2}: \dfrac{x-2}{3}=\dfrac{y-1}{4}=\dfrac{z-2}{5}$.

a⃗₁=(1,−1,1), b⃗₁=(2,3,4); a⃗₂=(2,1,2), b⃗₂=(3,4,5). a⃗₂−a⃗₁=(1,2,1).
b⃗₁×b⃗₂ = |i j k; 2 3 4; 3 4 5| = i(15−16)−j(10−12)+k(8−9) = (−1,2,−1).
|(a⃗₂−a⃗₁)·(b⃗₁×b⃗₂)| = |(1)(−1)+(2)(2)+(1)(−1)| = |−1+4−1| = 2. $|b⃗₁×b⃗₂| = \sqrt{1+4+1} = \sqrt6$.

✓ $SD = \dfrac{2}{\sqrt6} = 2\dfrac{\sqrt6}{6} =$ $\dfrac{\sqrt6}{3}$.

Show that the vectors $\vec a=(1,2,3),\ \vec b=(2,3,4),\ \vec c=(3,4,5)$ are coplanar, and interpret the result.

Vectors are coplanar iff the scalar triple product $[\vec a\ \vec b\ \vec c] = 0$, i.e. the determinant of their components vanishes.
$[\vec a\ \vec b\ \vec c] = \begin{vmatrix} 1 & 2 & 3 \\ 2 & 3 & 4 \\ 3 & 4 & 5 \end{vmatrix} = 1(15-16) - 2(10-12) + 3(8-9) = -1 + 4 - 3 = 0.$
Since the box product is $0$, the three vectors lie in one plane (here $\vec c = 2\vec b - \vec a$, a linear combination — confirming linear dependence).

✓ $[\vec a\ \vec b\ \vec c] = 0 \Rightarrow$ the vectors are coplanar.

Find the foot of the perpendicular from $P(2,3,4)$ to the plane $2x + y - z + 3 = 0$, and hence the distance $P$ to the plane.

Normal to the plane is $\vec n=(2,1,-1)$. Parametrize the perpendicular: $(x,y,z) = (2+2t,\ 3+t,\ 4-t).$
Substitute in the plane: $2(2+2t) + (3+t) - (4-t) + 3 = 0 \Rightarrow 4+4t+3+t-4+t+3 = 0 \Rightarrow 6t + 6 = 0 \Rightarrow t = -1.$
Foot $= (2-2,\ 3-1,\ 4+1) = (0,\ 2,\ 5).$ Distance $= |t|\,|\vec n| = 1\cdot\sqrt{4+1+1} = \sqrt6.$

✓ Foot $= (0,2,5)$; distance $= \sqrt6$.

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Vectors & 3D — Visual Concepts

Click any diagram to zoom

Dot Product, Cross Product, and Triple Product Geometry

Three key vector products. Dot product: scalar quantity = |a||b|cosθ, gives projection. Cross product: vector perpendicular to both, magnitude = area of parallelogram |a||b|sinθ. Scalar triple product [a b c] = a·(b×c) = volume of parallelepiped; zero iff vectors are coplanar.

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Vectors & 3D — Quick Quiz

Question 1

If a⃗·b⃗=0 and |a⃗|≠0 and |b⃗|≠0, then a⃗ and b⃗ are:

a⃗·b⃗ = |a||b|cosθ = 0. Since |a|≠0 and |b|≠0, cosθ = 0 → θ = 90° → vectors are perpendicular. For parallel: |a⃗×b⃗|=0 (cross product zero). Answer: B ✓

Question 2

The direction cosines of the line joining (1,2,3) and (4,−1,6) are proportional to:

Direction ratios = (4−1, −1−2, 6−3) = (3,−3,3) ∝ (1,−1,1). The direction cosines are $\dfrac{3,-3,3}{\sqrt{9+9+9}} = \dfrac{3,-3,3}{3\sqrt3} = (\dfrac{1}{\sqrt3}, -\dfrac{1}{\sqrt3}, \dfrac{1}{\sqrt3})$. Answer: C ✓

Question 3

Three vectors are coplanar if and only if their scalar triple product is:

[a b c] = a·(b×c) = volume of parallelepiped formed by a, b, c. Coplanar vectors → parallelepiped has zero volume → [a b c] = 0. This is the fundamental coplanarity condition. Answer: A ✓

Angle	sin	cos	tan	cosec	sec	cot
0°	0	1	0	—	1	—
30° (π/6)	\(\dfrac{1}{2}\)	\(\dfrac{\sqrt3}{2}\)	\(\dfrac{1}{\sqrt3}\)	\(2\)	\(\dfrac{2}{\sqrt3}\)	\(\sqrt3\)
45° (π/4)	\(\dfrac{1}{\sqrt2}\)	\(\dfrac{1}{\sqrt2}\)	\(1\)	\(\sqrt2\)	\(\sqrt2\)	\(1\)
60° (π/3)	\(\dfrac{\sqrt3}{2}\)	\(\dfrac{1}{2}\)	\(\sqrt3\)	\(\dfrac{2}{\sqrt3}\)	\(2\)	\(\dfrac{1}{\sqrt3}\)
90° (π/2)	1	0	—	1	—	0

Equation	General Solution
\(\sin\theta = \sin\alpha\)	\(\theta = n\pi + (-1)^{n}\alpha, n \in ℤ\)
\(\cos\theta = \cos\alpha\)	\(\theta = 2n\pi \pm \alpha, n \in ℤ\)
\(\tan\theta = \tan\alpha\)	\(\theta = n\pi + \alpha, n \in ℤ\)
\(\sin\theta = 0\)	\(\theta = n\pi\)
\(\cos\theta = 0\)	\(\theta = (2n+1)\dfrac{\pi}{2}\)
\(\tan\theta = 0\)	\(\theta = n\pi\)
\(\sin^{2}\theta = \sin^{2}\alpha\)	\(\theta = n\pi \pm \alpha\) (same as cos²θ = cos²α and tan²θ = tan²α)