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- Question 4 Exercise 8.2 @math-11-nbf:sol:unit08
- \cos 2 \theta$ %%(c)%% $\tan 2 \theta$ (d) $\sin \frac{\theta}{2}$ (e) $\cos \frac{\theta}{2}$ (f) $\tan \frac{\theta}{2}$ when: $\cos \theta=\frac{3}{5}$ where $0<\theta<\frac{\pi}{2}$ ** Solution. ** Given: $\c
- Question 1, Exercise 8.1 @math-11-nbf:sol:unit08
- \sin 60 \\ \implies \cos (180+60) & = (-1)\left(\frac{1}{2}\right) - (0)\left(\frac{\sqrt{3}}{2}\right) \\ & = -\frac{1}{2} - 0 = -\frac{1}{2} \end{align*} \begin{align*} \cos (\alpha - \beta) & = \cos \alpha \cos \be
- Exercise 1.1 (Solutions) @fsc-part1-ptb:sol:ch01
- mmutative property w.r.t. '+'. ---- (ii) $(a+1)+ \frac{3}{4}= a+(1+\frac{3}{4})$ **Property:** Associative property w.r.t. '+'. ---- (iii) $(\sqrt{3}+\sqrt{... cative property. ---- (v) $a>b \quad \Rightarrow \frac{1}{a}<\frac{1}{b}$. **Property:** Multiplicative property. ---- (vi) $a>b \quad \Rightarrow -a<-b$. *
- Exercise 6.2 @matric:9th_science
- ify as rational expression ====Question 1:==== $\frac{x^2-x-6}{x^2-9}+\frac{x^2+2x-24}{x^2-x-12}$\\ **Solution:**\\ $\begin{align} \frac{x^2-x-6}{x^2-9}&+\frac{x^2+2x-24}{x^2-x-12}\\ &=\frac{x^2-3x+2x-6}{(x)^2-(3)^2}+\frac{x^2+6x-4x-24}{x^2
- Exercise 2.6 (Solutions) @matric:9th_science:unit_02
- d write your answer in the form of $a+ib$ (i) $\frac{-2}{1+i}$\\ (ii) $\frac{2+3i}{4-i}$\\ (iii) $\frac{9-7i}{3+i}$\\ (iv) $\frac{2-6i}{3+i}-\frac{4+i}{3+i}$\\ (v) $({1+i}/{1-i})^2$\\ (vi) $\frac{1}{(2+3i)(1
- Question 3, Exercise 2.5 @math-11-nbf:sol:unit02
- nd{array} \right] \quad \text{by } R2-R_3 \quad \frac{1}{2} R3\\ \sim&{\text{R}} \left[ \begin{array}{c... 1 \\ 0 & 1 & 1 & -1 & 0 & 0 \\ 0 & 0 & 1 & 1 & \frac{1}{2} & \frac{1}{2} \end{array} \right] \quad \text{by } R1 + R2 \text{ and } \frac{1}{2} R3\\ \sim&{\text{R}} \left[ \begin{array}{c
- Question 5 Exercise 8.2 @math-11-nbf:sol:unit08
- eta$ using the information given: $\sin 2 \theta=\frac{24}{25}, 2 \theta$ in QII ** Solution. ** Given... = - \sqrt{1-\sin^2 2\theta}\\ &=- \sqrt{1-\left(\frac{24}{25}\right)^2} \\ &=- \sqrt{\frac{49}{625}} = -\frac{7}{25} \end{align*} Also we have $$\sin\theta = \pm \sqrt{\frac{1-\cos 2\theta}{2}}
- Question 12, Exercise 8.1 @math-11-nbf:sol:unit08
- +\beta) = \tan(180^{\circ}-\gamma) \\ \implies & \frac{\tan\alpha + \tan\beta}{1-\tan\alpha \tan\beta} ... pha+\beta+\gamma=180^{\circ}$, prove that: $\cot \frac{\alpha}{2}+\cot \frac{\beta}{2}+\cot \frac{\gamma}{2}=\cot \frac{\alpha}{2} \cot \frac{\beta}{2} \cot \frac{\gamma}{2}$ ** S
- Question 1 Exercise 7.3 @math-11-kpk:sol:unit07
- irst four terms in the expansion of (i) (1 - x) $\frac{1}{2}$ Solution: Using binomial theorem to tind the four terms $$ \begin{aligned} & (1-x)^{\frac{1}{2}}=1+\frac{1}{2} x+ \\ & \frac{\frac{1}{2}\left(-\frac{1}{2}-1\right)}{2 !}(-x)^2 \end{aligned} $$ $$ \begin{align
- Question 7 and 8, Exercise 4.8 @math-11-nbf:sol:unit04
- 7===== Find the sum of $n$ term of the series: $$\frac{1}{1 \times 4}+\frac{1}{4 \times 7}+\frac{1}{7 \times 10}+\ldots$$ ** Solution. ** Given: $$\frac{1}{1 \times 4}+\frac{1}{4 \times 7}+\frac{1}{7 \t
- Review exercise @matric:9th_science
- Quotient\\ **Answer:**\\ $c$\\ (xi) Simplify $\frac{a}{9a^2-b^2}+\frac{1}{3a-b}$= ---\\ (a) $\frac{4a}{9a^2-b^2}$ (b) $\frac{4a-b}{9a^2-b^2}$\\ (c) $\frac{4a+b}{9a^2-b^2}$ (d) $\frac{b}{9a
- Question 2, Review Exercise @math-11-nbf:sol:unit08
- n*} \cos^2 \theta &= 1-\sin^2\theta\\ &= 1-\left(\frac{3}{5}\right)^2\\ & =1-\frac{9}{25} \\ \implies \cos^2 \theta&=\frac{16}{25} \\ \cos \theta&=\pm\frac{4}{5}\\ \end{align*} As $\theta$ is obtuse, so $\theta$ lies in II
- Question 5 and 6 Exercise 7.3 @math-11-kpk:sol:unit07
- igher of $x$ may be negleeled. then show that $$ \frac{(8+3 x)^{\frac{2}{3}}}{(2+3 x) \sqrt{4-5 x}}=1-\frac{5 x}{8} $$ Solution: Given that: $$ \frac{\sqrt[4]{3}-3 x j^{\frac{2}{3}}}{2 \cdot 3 x+4-5 x} $$ $$ \b
- Question 2 Exercise 7.3 @math-11-kpk:sol:unit07
- & \sqrt{26}=\sqrt{25+1} \\ & =\sqrt{25} \sqrt{1+\frac{1}{25}}=5\left[1+\frac{1}{25}\right]^{\frac{1}{2}} \end{aligned} $$ Using binomial expansion $$ \begin{aligned} & \sqrt{26}=5\left[1+\frac{1}{25}\right]^{\frac{1}{2}} \\ & =5\left[1+\frac{
- Question 4, Exercise 2.6 @math-11-nbf:sol:unit02
- nd{bmatrix}\\ &\sim \text{R}\begin{bmatrix} 1 & -\frac{1}{2} & -\frac{1}{2} & \vert & 1 \\ 3 & -4 & 3 & \vert & 7 \\ 4 & 2 & -5 & \vert & 10 \end{bmatrix}\qua... dfrac{1}{2}\\ &\sim \text{R}\begin{bmatrix} 1 & -\frac{1}{2} & -\frac{1}{2} & : & 1 \\ 0 & -\frac{5}{2} & \frac{9}{2} & : & 4 \\ 4 & 2 & -5 & : & 10 \end{bmat