Maharashtra HSC Board Maths Theorems

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1. Napier's Analogy (Triangle ABC) In $\triangle ABC$, the following relations hold: $\tan\left(\frac{B-C}{2}\right) = \frac{b-c}{b+c} \cot\left(\frac{A}{2}\right)$ $\tan\left(\frac{C-A}{2}\right) = \frac{c-a}{c+a} \cot\left(\frac{B}{2}\right)$ $\tan\left(\frac{A-B}{2}\right) = \frac{a-b}{a+b} \cot\left(\frac{C}{2}\right)$ Proof of $\tan\left(\frac{B-C}{2}\right) = \frac{b-c}{b+c} \cot\left(\frac{A}{2}\right)$ By Sine Rule, $b = 2R\sin B$ and $c = 2R\sin C$. $\frac{b-c}{b+c} = \frac{2R\sin B - 2R\sin C}{2R\sin B + 2R\sin C} = \frac{\sin B - \sin C}{\sin B + \sin C}$ Using sum-to-product formulas: $\sin B - \sin C = 2\cos\left(\frac{B+C}{2}\right)\sin\left(\frac{B-C}{2}\right)$ $\sin B + \sin C = 2\sin\left(\frac{B+C}{2}\right)\cos\left(\frac{B-C}{2}\right)$ So, $\frac{b-c}{b+c} = \frac{2\cos\left(\frac{B+C}{2}\right)\sin\left(\frac{B-C}{2}\right)}{2\sin\left(\frac{B+C}{2}\right)\cos\left(\frac{B-C}{2}\right)} = \cot\left(\frac{B+C}{2}\right)\tan\left(\frac{B-C}{2}\right)$ In a triangle, $A+B+C = \pi$, so $\frac{B+C}{2} = \frac{\pi}{2} - \frac{A}{2}$. Thus, $\cot\left(\frac{B+C}{2}\right) = \cot\left(\frac{\pi}{2} - \frac{A}{2}\right) = \tan\left(\frac{A}{2}\right)$. Substituting this back: $\frac{b-c}{b+c} = \tan\left(\frac{A}{2}\right)\tan\left(\frac{B-C}{2}\right)$ Therefore, $\tan\left(\frac{B-C}{2}\right) = \frac{b-c}{b+c} \cot\left(\frac{A}{2}\right)$. 2. Half-Angle Formulas (Triangle ABC) Let $s = \frac{a+b+c}{2}$ be the semi-perimeter. (i) Sine Formula $\sin\left(\frac{A}{2}\right) = \sqrt{\frac{(s-b)(s-c)}{bc}}$ Proof: We use the cosine rule: $\cos A = \frac{b^2+c^2-a^2}{2bc}$. We know $1 - \cos A = 2\sin^2\left(\frac{A}{2}\right)$. $2\sin^2\left(\frac{A}{2}\right) = 1 - \frac{b^2+c^2-a^2}{2bc} = \frac{2bc - b^2 - c^2 + a^2}{2bc} = \frac{a^2 - (b^2 - 2bc + c^2)}{2bc} = \frac{a^2 - (b-c)^2}{2bc}$ $2\sin^2\left(\frac{A}{2}\right) = \frac{(a-(b-c))(a+(b-c))}{2bc} = \frac{(a-b+c)(a+b-c)}{2bc}$ Since $a+b+c=2s$, we have: $a-b+c = (a+c+b) - 2b = 2s - 2b = 2(s-b)$ $a+b-c = (a+b+c) - 2c = 2s - 2c = 2(s-c)$ So, $2\sin^2\left(\frac{A}{2}\right) = \frac{2(s-b)2(s-c)}{2bc} = \frac{2(s-b)(s-c)}{bc}$ $\sin^2\left(\frac{A}{2}\right) = \frac{(s-b)(s-c)}{bc}$ $\sin\left(\frac{A}{2}\right) = \sqrt{\frac{(s-b)(s-c)}{bc}}$ (Since $A/2$ is acute, $\sin(A/2) > 0$) (ii) Cosine Formula $\cos\left(\frac{A}{2}\right) = \sqrt{\frac{s(s-a)}{bc}}$ Proof: We know $1 + \cos A = 2\cos^2\left(\frac{A}{2}\right)$. $2\cos^2\left(\frac{A}{2}\right) = 1 + \frac{b^2+c^2-a^2}{2bc} = \frac{2bc + b^2 + c^2 - a^2}{2bc} = \frac{(b+c)^2 - a^2}{2bc}$ $2\cos^2\left(\frac{A}{2}\right) = \frac{(b+c-a)(b+c+a)}{2bc}$ Since $a+b+c=2s$, we have: $b+c-a = (a+b+c) - 2a = 2s - 2a = 2(s-a)$ $b+c+a = 2s$ So, $2\cos^2\left(\frac{A}{2}\right) = \frac{2(s-a)2s}{2bc} = \frac{2s(s-a)}{bc}$ $\cos^2\left(\frac{A}{2}\right) = \frac{s(s-a)}{bc}$ $\cos\left(\frac{A}{2}\right) = \sqrt{\frac{s(s-a)}{bc}}$ (Since $A/2$ is acute, $\cos(A/2) > 0$) (iii) Tangent Formula $\tan\left(\frac{A}{2}\right) = \sqrt{\frac{(s-b)(s-c)}{s(s-a)}}$ Proof: $\tan\left(\frac{A}{2}\right) = \frac{\sin(A/2)}{\cos(A/2)} = \frac{\sqrt{\frac{(s-b)(s-c)}{bc}}}{\sqrt{\frac{s(s-a)}{bc}}} = \sqrt{\frac{(s-b)(s-c)}{s(s-a)}}$ 3. Coplanarity of Vectors Theorem: If $\vec{a}$ and $\vec{b}$ are two non-zero and non-collinear vectors, then any vector $\vec{r}$ coplanar with $\vec{a}$ and $\vec{b}$ can be expressed as $\vec{r} = t_1\vec{a} + t_2\vec{b}$ where $t_1$ and $t_2$ are unique scalars. Proof (Only If-part: Existence) Let $\vec{r}$ be coplanar with $\vec{a}$ and $\vec{b}$. Let $O$ be the origin. Let $\vec{OA} = \vec{a}$, $\vec{OB} = \vec{b}$, and $\vec{OP} = \vec{r}$. Draw a line through $P$ parallel to $OB$ meeting $OA$ at $M$. Draw a line through $P$ parallel to $OA$ meeting $OB$ at $N$. By vector addition (triangle or parallelogram law), $\vec{OP} = \vec{OM} + \vec{ON}$. Since $M$ lies on $OA$ and $N$ lies on $OB$, $\vec{OM}$ must be a scalar multiple of $\vec{a}$, say $\vec{OM} = t_1\vec{a}$. Similarly, $\vec{ON}$ must be a scalar multiple of $\vec{b}$, say $\vec{ON} = t_2\vec{b}$. Thus, $\vec{r} = t_1\vec{a} + t_2\vec{b}$ for some scalars $t_1, t_2 \in \mathbb{R}$. Proof (If-part: Coplanarity) If $\vec{r} = t_1\vec{a} + t_2\vec{b}$, then $\vec{r}$ is a linear combination of $\vec{a}$ and $\vec{b}$. Since $\vec{a}$ and $\vec{b}$ are coplanar, any linear combination of $\vec{a}$ and $\vec{b}$ must also lie in the same plane. Hence, $\vec{r}$, $\vec{a}$, and $\vec{b}$ are coplanar. Proof (Uniqueness) Assume $\vec{r}$ can be expressed in two ways: $\vec{r} = t_1\vec{a} + t_2\vec{b}$ (1) $\vec{r} = s_1\vec{a} + s_2\vec{b}$ (2) Subtracting (2) from (1): $\vec{0} = (t_1-s_1)\vec{a} + (t_2-s_2)\vec{b}$ Since $\vec{a}$ and $\vec{b}$ are non-collinear vectors, their linear combination can only be zero if the coefficients are both zero. Thus, $t_1-s_1 = 0 \Rightarrow t_1 = s_1$ And $t_2-s_2 = 0 \Rightarrow t_2 = s_2$ This proves the uniqueness of the scalars $t_1$ and $t_2$. 4. Condition for Coplanarity of Three Vectors Theorem: Three vectors $\vec{a}, \vec{b}, \vec{c}$ are coplanar if and only if there exists a non-zero linear combination $x\vec{a} + y\vec{b} + z\vec{c} = \vec{0}$ with $(x,y,z) \neq (0,0,0)$. Proof (Only If-part: Existence of Non-zero Linear Combination) Assume $\vec{a}, \vec{b}, \vec{c}$ are coplanar. Case 1: If any two vectors are collinear (e.g., $\vec{a}$ and $\vec{b}$). Then $\vec{a} = k\vec{b}$ for some scalar $k \neq 0$. So, $\vec{a} - k\vec{b} + 0\vec{c} = \vec{0}$. Here, $(x,y,z) = (1, -k, 0)$, which is non-zero, satisfying the condition. Case 2: If no two vectors are collinear. Since $\vec{a}, \vec{b}, \vec{c}$ are coplanar and $\vec{a}, \vec{b}$ are non-collinear, by the previous theorem, $\vec{c}$ can be expressed as a linear combination of $\vec{a}$ and $\vec{b}$. So, $\vec{c} = x\vec{a} + y\vec{b}$ for some scalars $x, y$. Rearranging, $x\vec{a} + y\vec{b} - \vec{c} = \vec{0}$. Here, $(x,y,z) = (x, y, -1)$, which is non-zero, satisfying the condition. Proof (If-part: Coplanarity of Vectors) Assume there exist scalars $x, y, z$, not all zero, such that $x\vec{a} + y\vec{b} + z\vec{c} = \vec{0}$. Without loss of generality, let $z \neq 0$. Then $z\vec{c} = -x\vec{a} - y\vec{b}$ $\vec{c} = \left(-\frac{x}{z}\right)\vec{a} + \left(-\frac{y}{z}\right)\vec{b}$ This shows that $\vec{c}$ is a linear combination of $\vec{a}$ and $\vec{b}$. Therefore, $\vec{c}$ lies in the plane formed by $\vec{a}$ and $\vec{b}$. Hence, $\vec{a}, \vec{b}, \vec{c}$ are coplanar. 5. Condition for Collinearity of Two Vectors Theorem: Two non-zero vectors $\vec{a}$ and $\vec{b}$ are collinear if and only if there exist scalars $m$ and $n$, not both zero, such that $m\vec{a} + n\vec{b} = \vec{0}$. Proof (Only If-part: Existence of Non-zero Linear Combination) Assume $\vec{a}$ and $\vec{b}$ are collinear. Since they are non-zero and collinear, one must be a scalar multiple of the other. Let $\vec{a} = t\vec{b}$ for some scalar $t \neq 0$. Then $\vec{a} - t\vec{b} = \vec{0}$. Here, $m=1$ and $n=-t$, which are not both zero, satisfying the condition. Proof (If-part: Collinearity of Vectors) Assume there exist scalars $m, n$, not both zero, such that $m\vec{a} + n\vec{b} = \vec{0}$. Without loss of generality, let $m \neq 0$. Then $m\vec{a} = -n\vec{b}$ $\vec{a} = \left(-\frac{n}{m}\right)\vec{b}$ Let $k = -\frac{n}{m}$. Then $\vec{a} = k\vec{b}$. This means $\vec{a}$ is a scalar multiple of $\vec{b}$. Since $\vec{a}$ and $\vec{b}$ are non-zero, they must be collinear. 6. Angle Subtended by a Semicircle is a Right Angle (Vector Method) Theorem: The angle subtended by a semicircle at any point on the circumference is a right angle. Proof Let $AB$ be the diameter of a circle with center $C$. Let $P$ be any point on the circumference. We need to prove $\angle APB = 90^\circ$, which means $\vec{PA} \cdot \vec{PB} = 0$. Let $C$ be the origin. Then $\vec{CA}$ and $\vec{CB}$ are radius vectors. Since $AB$ is a diameter, $\vec{CB} = -\vec{CA}$. Let $\vec{CA} = \vec{a}$. Then $\vec{CB} = -\vec{a}$. Let $\vec{CP} = \vec{r}$. Since $P$ is on the circle, $|\vec{CP}| = |\vec{CA}| = |\vec{CB}|$, so $|\vec{r}| = |\vec{a}|$. This implies $\vec{r} \cdot \vec{r} = \vec{a} \cdot \vec{a}$. Now, consider vectors $\vec{PA}$ and $\vec{PB}$: $\vec{PA} = \vec{CA} - \vec{CP} = \vec{a} - \vec{r}$ $\vec{PB} = \vec{CB} - \vec{CP} = -\vec{a} - \vec{r}$ Calculate their dot product: $\vec{PA} \cdot \vec{PB} = (\vec{a} - \vec{r}) \cdot (-\vec{a} - \vec{r})$ $\vec{PA} \cdot \vec{PB} = -(\vec{a} - \vec{r}) \cdot (\vec{a} + \vec{r})$ $\vec{PA} \cdot \vec{PB} = -(\vec{a} \cdot \vec{a} + \vec{a} \cdot \vec{r} - \vec{r} \cdot \vec{a} - \vec{r} \cdot \vec{r})$ Since dot product is commutative ($\vec{a} \cdot \vec{r} = \vec{r} \cdot \vec{a}$): $\vec{PA} \cdot \vec{PB} = -(\vec{a} \cdot \vec{a} - \vec{r} \cdot \vec{r})$ As established, $\vec{a} \cdot \vec{a} = \vec{r} \cdot \vec{r}$ (because $|\vec{a}| = |\vec{r}|$). So, $\vec{PA} \cdot \vec{PB} = -(\vec{a} \cdot \vec{a} - \vec{a} \cdot \vec{a}) = 0$. Since the dot product is zero, $\vec{PA}$ is perpendicular to $\vec{PB}$. Therefore, $\angle APB = 90^\circ$. 7. Angle between Lines Represented by $ax^2 + 2hxy + by^2 = 0$ Theorem: If $\theta$ is the acute angle between the lines represented by $ax^2 + 2hxy + by^2 = 0$, then $\tan\theta = \left|\frac{2\sqrt{h^2-ab}}{a+b}\right|$. Proof The homogeneous equation $ax^2 + 2hxy + by^2 = 0$ represents a pair of straight lines passing through the origin. Let their slopes be $m_1$ and $m_2$. Dividing the equation by $x^2$ (assuming $x \neq 0$): $a + 2h\left(\frac{y}{x}\right) + b\left(\frac{y}{x}\right)^2 = 0$ $b\left(\frac{y}{x}\right)^2 + 2h\left(\frac{y}{x}\right) + a = 0$ Replacing $y/x$ with $m$ (slope): $bm^2 + 2hm + a = 0$. This is a quadratic equation in $m$, whose roots are $m_1$ and $m_2$. From Vieta's formulas: Sum of roots: $m_1 + m_2 = -\frac{2h}{b}$ Product of roots: $m_1 m_2 = \frac{a}{b}$ The formula for the angle $\theta$ between two lines with slopes $m_1$ and $m_2$ is $\tan\theta = \left|\frac{m_1-m_2}{1+m_1m_2}\right|$. First, find $(m_1-m_2)^2$: $(m_1-m_2)^2 = (m_1+m_2)^2 - 4m_1m_2$ $(m_1-m_2)^2 = \left(-\frac{2h}{b}\right)^2 - 4\left(\frac{a}{b}\right) = \frac{4h^2}{b^2} - \frac{4a}{b} = \frac{4h^2 - 4ab}{b^2} = \frac{4(h^2-ab)}{b^2}$ So, $|m_1-m_2| = \sqrt{\frac{4(h^2-ab)}{b^2}} = \frac{2\sqrt{h^2-ab}}{|b|}$. Now substitute into the $\tan\theta$ formula: $\tan\theta = \left|\frac{\frac{2\sqrt{h^2-ab}}{|b|}}{1+\frac{a}{b}}\right| = \left|\frac{\frac{2\sqrt{h^2-ab}}{|b|}}{\frac{b+a}{b}}\right|$ If $b>0$: $\tan\theta = \left|\frac{\frac{2\sqrt{h^2-ab}}{b}}{\frac{a+b}{b}}\right| = \left|\frac{2\sqrt{h^2-ab}}{a+b}\right|$ If $b Thus, for any $b \neq 0$, $\tan\theta = \left|\frac{2\sqrt{h^2-ab}}{a+b}\right|$. Special cases: If $h^2-ab > 0$, the lines are real and distinct. If $h^2-ab = 0$, the lines are real and coincident ($\theta = 0$). If $h^2-ab If $a+b = 0$, the lines are perpendicular ($\tan\theta$ is undefined, $\theta = 90^\circ$).