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Exercise 1.2 (Solutions) @fsc-part1-ptb:sol:ch01
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iv (3,7)$ **Solutions** $\quad (2,6)\div (3,7)=\dfrac{(2,6)}{\left( 3,7 \right)}$ $=\dfrac{2+6i}{3+7i}=\dfrac{2+6i}{3+7i}\times \dfrac{3-7i}{3-7i}$ $=\dfrac{6-14i+18i-42{{i}^{2}}}{{{\left( 3 \right)}^{2}}-{{\left(
Solution and Area of Oblique Triangle
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\cos \gamma$ </col><col sm="6"> * $\cos\alpha =\dfrac{b^2+c^2-a^2}{2bc}$ * $\cos\beta =\dfrac{c^2+a^2-b^2}{2ac}$ * $\cos\gamma =\dfrac{a^2+b^2-c^2}{2ab}$ </col></grid> </panel> <panel title="The Law of Sine"> * $\dfrac{a}{\sin \alpha }=\dfrac{b}{\sin \beta }=\dfrac{c}
Trigonometric Formulas
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sm="6"> * $\tan \left( \alpha +\beta \right)=\dfrac{\tan \alpha +\tan \beta }{1-\tan \alpha \tan \beta }$ * $\tan \left( \alpha -\beta \right)=\dfrac{\tan \alpha -\tan \beta }{1+\tan \alpha \tan \bet... theta -{{\sin }^{2}}\theta$ * $\tan 2\theta =\dfrac{2\tan \theta }{1-{{\tan }^{2}}\theta }$ </col><col sm="6"> * ${{\sin }^{2}}\frac{\theta }{2}=\dfrac{1-\cos \theta }{2}$ * ${{\cos }^{2}}\frac{\thet
Exercise 1.1 (Solutions) @fsc-part1-ptb:sol:ch01
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step: $$\frac{4+16x}{4}$$ **Solution** $\quad \dfrac{4+16x}{4}$ $=\dfrac{1}{4}\times (4+16x)\quad \because\dfrac{a}{b}=\dfrac{1}{b}\times a$ $=\dfrac{1}{4} \times (4\times 1+4 \times 4x)\quad \text{(multiplicative id
Definitions: FSc Part 1 (Mathematics): PTB by Aurang Zaib
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= A number which can be expressed in the form \( \dfrac{p}{q} \), where \( p, q \in \mathbb{Z} \) and \( ... s termed as a rational number. ===Example==== \( \dfrac{3}{4} \), \( \dfrac{7}{2} \) ====Irrational Number==== A real number that cannot be represented as a fr... ers include \( 2 \), \( 3 + \sqrt{3}i \), and \( \dfrac{1}{2} + i \). ====Real Plane or Coordinate Plane
Exercise 2.8 (Solutions) @fsc-part1-ptb:sol:ch02
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ity element. d- For $a\in {{\mathbb{Q}}^{+}}$, $\dfrac{1}{a}\in {{\mathbb{Q}}^{+}}$ such that $a\times \dfrac{1}{a}=\dfrac{1}{a}\times a=1$. Thus inverse of $a$ is $\dfrac{1}{a}$. Hence ${{\mathbb{Q}}^{+}}$ is group under addi