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: $c \cos B + b \cos C = c(0) + b \cos C = b \cos C = a = \text{LHS}$.

HSC Physics Board Papers with Solution

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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 $x = a \tan \theta \Rightarrow \theta = \tan^{-1}(x/a)$: $$ I = \frac{1}{a} \tan^{-1}\left(\frac{x}{a}\right) + c $$