HSC Maths Theorems & Proofs
Based on New Syllabus 2020 - Important Chapters
Mathematics-I
3. Trigonometric Functions
Sine Rule: In \(\Delta ABC\), prove that \(\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = 2R\), where R is the circumradius.
Proof
Let Area of \(\Delta ABC\) be denoted by \(A(\Delta ABC)\).
(i) We know that area of triangle is half the product of two sides and the sine of the included angle.
$$ A(\Delta ABC) = \frac{1}{2} bc \sin A = \frac{1}{2} ac \sin B = \frac{1}{2} ab \sin C $$
(ii) Multiply throughout by 2:
$$ 2 A(\Delta ABC) = bc \sin A = ac \sin B = ab \sin C $$
(iii) Divide throughout by \(abc\):
$$ \frac{bc \sin A}{abc} = \frac{ac \sin B}{abc} = \frac{ab \sin C}{abc} $$
$$ \therefore \frac{\sin A}{a} = \frac{\sin B}{b} = \frac{\sin C}{c} = k \quad \dots \text{(constant)} $$
Taking reciprocals:
$$ \frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} \quad \dots (1) $$
(iv) Relation with R (Circumradius):
Consider the circumcircle of \(\Delta ABC\) with center O and radius R.
Draw diameter AP. Length \(AP = 2R\). Join P to C.
In \(\Delta ACP\), \(\angle ACP = 90^\circ\) (Angle in a semicircle).
Also, \(\angle ABC = \angle APC\) (Angles inscribed in the same arc).
$$ \therefore \angle B = \angle P $$
In right-angled \(\Delta ACP\):
$$ \sin P = \frac{\text{Opposite}}{\text{Hypotenuse}} = \frac{AC}{AP} = \frac{b}{2R} $$
Since \(\angle B = \angle P\), \(\sin B = \frac{b}{2R}\).
$$ \therefore \frac{b}{\sin B} = 2R $$
From (1), we get:
$$ \frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = 2R $$
12th English Board Papers
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Cosine Rule: In \(\Delta ABC\), with usual notations, prove that \(b^2 = c^2 + a^2 - 2ca \cos B\).
Proof
(i) Let \(\Delta ABC\) be placed in the Cartesian coordinate system such that vertex B is at the origin \((0,0)\). Side BC lies along the positive X-axis.
The coordinates of the vertices are:
$$ B \equiv (0,0) $$
$$ C \equiv (a, 0) $$
$$ A \equiv (c \cos B, c \sin B) $$
(ii) Using the distance formula for length \(b = l(AC)\):
$$ b^2 = (x_2 - x_1)^2 + (y_2 - y_1)^2 $$
$$ b^2 = (c \cos B - a)^2 + (c \sin B - 0)^2 $$
(iii) Expand the squares:
$$ b^2 = (c^2 \cos^2 B - 2ac \cos B + a^2) + c^2 \sin^2 B $$
$$ b^2 = c^2 (\cos^2 B + \sin^2 B) + a^2 - 2ac \cos B $$
(iv) Since \(\sin^2 B + \cos^2 B = 1\):
$$ b^2 = c^2(1) + a^2 - 2ac \cos B $$
$$ b^2 = c^2 + a^2 - 2ac \cos B $$
Hence proved.
Projection Rule: In \(\Delta ABC\), prove that \(a = c \cos B + b \cos C\).
Proof
Draw altitude AD from vertex A perpendicular to side BC (or BC produced). Let D be the foot of the perpendicular.
Case (i): B and C are acute angles
In right-angled \(\Delta ADB\):
$$ \cos B = \frac{BD}{AB} \Rightarrow BD = AB \cos B = c \cos B $$
In right-angled \(\Delta ADC\):
$$ \cos C = \frac{DC}{AC} \Rightarrow DC = AC \cos C = b \cos C $$
From the figure, \(BC = BD + DC\):
$$ a = c \cos B + b \cos C $$
Case (ii): Angle B is obtuse
Point D lies on BC produced to the left.
In \(\Delta ABD\), \(\angle ABD = 180^\circ - B\) (linear pair).
$$ \cos(180^\circ - B) = \frac{BD}{c} $$
$$ -\cos B = \frac{BD}{c} \Rightarrow BD = -c \cos B $$
In \(\Delta ADC\):
$$ \cos C = \frac{DC}{b} \Rightarrow DC = b \cos C $$
From the figure, \(BC = DC - DB\):
$$ a = b \cos C - (-c \cos B) = b \cos C + c \cos B $$
Case (iii): Angle B is a right angle (\(90^\circ\))
$$ \cos B = \cos 90^\circ = 0 $$
In \(\Delta ABC\), \(\cos C = \frac{BC}{AC} = \frac{a}{b} \Rightarrow a = b \cos C\).
RHS: \(c \cos B + b \cos C = c(0) + b \cos C = b \cos C = a = \text{LHS}\).
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4. Pair of Straight Lines
Theorem: The homogeneous equation of degree two in \(x\) and \(y\), \(ax^2 + 2hxy + by^2 = 0\), represents a pair of lines passing through the origin if \(h^2 - ab \geqslant 0\).
Proof
Consider the equation \(ax^2 + 2hxy + by^2 = 0\).
Case (i): If \(b = 0\)
The equation becomes \(ax^2 + 2hxy = 0\).
$$ x(ax + 2hy) = 0 $$
This represents two lines: \(x = 0\) (Y-axis) and \(ax + 2hy = 0\). Both pass through the origin.
Case (ii): If \(b \neq 0\)
Multiply the equation by \(b\):
$$ abx^2 + 2hbxy + b^2y^2 = 0 $$
Rearrange to complete the square:
$$ b^2y^2 + 2hbxy = -abx^2 $$
Add \(h^2x^2\) to both sides:
$$ b^2y^2 + 2hbxy + h^2x^2 = h^2x^2 - abx^2 $$
$$ (by + hx)^2 = x^2(h^2 - ab) $$
$$ (by + hx)^2 = (x\sqrt{h^2 - ab})^2 \quad (\because h^2 - ab \geqslant 0) $$
Taking square roots:
$$ by + hx = \pm x\sqrt{h^2 - ab} $$
$$ by = -hx \pm x\sqrt{h^2 - ab} $$
$$ y = \left( \frac{-h \pm \sqrt{h^2 - ab}}{b} \right)x $$
This represents two lines passing through the origin, \(y = m_1x\) and \(y = m_2x\).
Theorem: The acute angle \(\theta\) between the lines represented by \(ax^2 + 2hxy + by^2 = 0\) is given by \(\tan \theta = \left| \frac{2\sqrt{h^2 - ab}}{a + b} \right|\).
Proof
Let \(m_1\) and \(m_2\) be the slopes of the lines.
From the theory of quadratic equations:
$$ m_1 + m_2 = -\frac{2h}{b} \quad \text{and} \quad m_1m_2 = \frac{a}{b} $$
We know that \(\tan \theta = \left| \frac{m_1 - m_2}{1 + m_1m_2} \right|\).
First, calculate \((m_1 - m_2)^2\):
$$ (m_1 - m_2)^2 = (m_1 + m_2)^2 - 4m_1m_2 $$
$$ = \left(-\frac{2h}{b}\right)^2 - 4\left(\frac{a}{b}\right) $$
$$ = \frac{4h^2}{b^2} - \frac{4a}{b} = \frac{4h^2 - 4ab}{b^2} $$
$$ \therefore m_1 - m_2 = \pm \frac{2\sqrt{h^2 - ab}}{b} $$
Substitute in the formula for \(\tan \theta\):
$$ \tan \theta = \left| \frac{\frac{2\sqrt{h^2 - ab}}{b}}{1 + \frac{a}{b}} \right| $$
$$ \tan \theta = \left| \frac{\frac{2\sqrt{h^2 - ab}}{b}}{\frac{b + a}{b}} \right| $$
$$ \tan \theta = \left| \frac{2\sqrt{h^2 - ab}}{a + b} \right| $$
Hence proved.
5. Vectors
Theorem: Two non-zero vectors \(\bar{a}\) and \(\bar{b}\) are collinear if and only if there exist scalars \(m\) and \(n\), at least one of them non-zero, such that \(m\bar{a} + n\bar{b} = \bar{0}\).
Proof
(i) Assume \(\bar{a}\) and \(\bar{b}\) are collinear.
Then \(\bar{a} = t\bar{b}\) for some scalar \(t \neq 0\).
$$ \bar{a} - t\bar{b} = \bar{0} $$
Let \(m = 1\) and \(n = -t\).
Thus, \(m\bar{a} + n\bar{b} = \bar{0}\) where \(m \neq 0\).
(ii) Conversely, assume \(m\bar{a} + n\bar{b} = \bar{0}\) and \(m \neq 0\).
$$ m\bar{a} = -n\bar{b} $$
$$ \bar{a} = \left(-\frac{n}{m}\right)\bar{b} $$
Since \(\bar{a}\) is a scalar multiple of \(\bar{b}\), the vectors are collinear.
Theorem: Let \(\bar{a}\) and \(\bar{b}\) be non-collinear vectors. A vector \(\bar{r}\) is coplanar with them if and only if \(\bar{r} = t_1\bar{a} + t_2\bar{b}\) uniquely.
Proof
Existence: Let \(\bar{a}\) be along OA and \(\bar{b}\) be along OB. Let \(\bar{r}\) be along OP.
Complete the parallelogram OMPN with OP as diagonal, \(M\) on OA, \(N\) on OB.
By parallelogram law: \(\vec{OP} = \vec{OM} + \vec{ON}\).
Since \(\vec{OM}\) is collinear with \(\bar{a}\), \(\vec{OM} = t_1\bar{a}\).
Since \(\vec{ON}\) is collinear with \(\bar{b}\), \(\vec{ON} = t_2\bar{b}\).
$$ \therefore \bar{r} = t_1\bar{a} + t_2\bar{b} $$
Uniqueness: Suppose \(\bar{r} = t_1\bar{a} + t_2\bar{b}\) and also \(\bar{r} = s_1\bar{a} + s_2\bar{b}\).
Subtracting:
$$ \bar{0} = (t_1 - s_1)\bar{a} + (t_2 - s_2)\bar{b} $$
Since \(\bar{a}\) and \(\bar{b}\) are non-collinear, the scalars must be zero.
$$ t_1 - s_1 = 0 \Rightarrow t_1 = s_1 $$
$$ t_2 - s_2 = 0 \Rightarrow t_2 = s_2 $$
Hence the representation is unique.
Section Formula (Internal Division): If \(R(\bar{r})\) divides segment AB joining \(A(\bar{a})\) and \(B(\bar{b})\) internally in ratio \(m:n\), then \(\bar{r} = \frac{m\bar{b} + n\bar{a}}{m + n}\).
Proof
Since R divides AB internally in ratio \(m:n\), A-R-B are collinear and \(\frac{AR}{RB} = \frac{m}{n}\).
$$ n(AR) = m(RB) $$
Since direction is same (A to R and R to B):
$$ n(\vec{AR}) = m(\vec{RB}) $$
Using position vectors relative to origin O:
$$ n(\bar{r} - \bar{a}) = m(\bar{b} - \bar{r}) $$
$$ n\bar{r} - n\bar{a} = m\bar{b} - m\bar{r} $$
$$ \bar{r}(m + n) = m\bar{b} + n\bar{a} $$
$$ \bar{r} = \frac{m\bar{b} + n\bar{a}}{m + n} $$
Mathematics-II
1. Differentiation
Chain Rule: If \(y = f(u)\) is a differentiable function of \(u\) and \(u = g(x)\) is a differentiable function of \(x\), then \(\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}\).
Proof
Let \(\delta x\) be a small increment in \(x\).
Let \(\delta u\) and \(\delta y\) be the corresponding increments in \(u\) and \(y\).
As \(\delta x \to 0\), \(\delta u \to 0\) and \(\delta y \to 0\).
Consider the increment ratio:
$$ \frac{\delta y}{\delta x} = \frac{\delta y}{\delta u} \times \frac{\delta u}{\delta x} \quad (\text{assuming } \delta u \neq 0) $$
Taking limits on both sides as \(\delta x \to 0\):
$$ \lim_{\delta x \to 0} \frac{\delta y}{\delta x} = \lim_{\delta x \to 0} \left( \frac{\delta y}{\delta u} \times \frac{\delta u}{\delta x} \right) $$
$$ \lim_{\delta x \to 0} \frac{\delta y}{\delta x} = \left(\lim_{\delta u \to 0} \frac{\delta y}{\delta u}\right) \times \left(\lim_{\delta x \to 0} \frac{\delta u}{\delta x}\right) $$
Since \(y\) is differentiable w.r.t \(u\) and \(u\) w.r.t \(x\), the limits exist and are equal to derivatives:
$$ \frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} $$
Parametric Function: If \(x = f(t)\) and \(y = g(t)\) are differentiable functions of \(t\), then \(\frac{dy}{dx} = \frac{dy/dt}{dx/dt}\), provided \(\frac{dx}{dt} \neq 0\).
Proof
Let \(\delta t\) be a small increment in \(t\).
Let \(\delta x\) and \(\delta y\) be corresponding increments in \(x\) and \(y\).
Consider the ratio:
$$ \frac{\delta y}{\delta x} = \frac{\delta y / \delta t}{\delta x / \delta t} \quad (\delta x \neq 0) $$
Taking limit as \(\delta t \to 0\) (implies \(\delta x \to 0\)):
$$ \lim_{\delta x \to 0} \frac{\delta y}{\delta x} = \frac{\lim_{\delta t \to 0} (\delta y / \delta t)}{\lim_{\delta t \to 0} (\delta x / \delta t)} $$
$$ \frac{dy}{dx} = \frac{dy/dt}{dx/dt} $$
HSC Chemistry
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3. Indefinite Integration
Integration by Parts: If \(u\) and \(v\) are differentiable functions of \(x\), then \(\int uv \, dx = u \int v \, dx - \int \left( \frac{du}{dx} \cdot \int v \, dx \right) dx\).
Proof
Let \(\int v \, dx = w\). Then \(\frac{dw}{dx} = v\).
Consider the derivative of the product \(u \cdot w\):
$$ \frac{d}{dx}(uw) = u \frac{dw}{dx} + w \frac{du}{dx} $$
Substitute \(\frac{dw}{dx} = v\):
$$ \frac{d}{dx}(uw) = uv + w \frac{du}{dx} $$
Rearranging terms:
$$ uv = \frac{d}{dx}(uw) - w \frac{du}{dx} $$
Integrating both sides w.r.t \(x\):
$$ \int uv \, dx = uw - \int \left( w \frac{du}{dx} \right) dx $$
Substitute back \(w = \int v \, dx\):
$$ \int uv \, dx = u \int v \, dx - \int \left( \frac{du}{dx} \int v \, dx \right) dx $$
Hence proved.
Special Integral: Prove that \(\int \frac{1}{x^2 + a^2} dx = \frac{1}{a} \tan^{-1}\left(\frac{x}{a}\right) + c\).
Proof
Put \(x = a \tan \theta\).
Differentiating w.r.t \(\theta\):
$$ dx = a \sec^2 \theta \, d\theta $$
Substitute in the integral:
$$ I = \int \frac{1}{a^2 \tan^2 \theta + a^2} \cdot a \sec^2 \theta \, d\theta $$
$$ I = \int \frac{a \sec^2 \theta}{a^2(\tan^2 \theta + 1)} \, d\theta $$
We know \(1 + \tan^2 \theta = \sec^2 \theta\):
$$ I = \int \frac{a \sec^2 \theta}{a^2 \sec^2 \theta} \, d\theta $$
$$ I = \frac{1}{a} \int 1 \, d\theta = \frac{1}{a} \theta + c $$
Since \(x = a \tan \theta \Rightarrow \theta = \tan^{-1}(x/a)\):
$$ I = \frac{1}{a} \tan^{-1}\left(\frac{x}{a}\right) + c $$