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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
- PPSC Paper 2015 (Lecturer in Mathematics) @ppsc
- ng this paper. - \(\displaystyle \int_{-4}^{0}\frac{tdt}{\sqrt{16-t62}}\) - $0$ - divergent ... the function\(A\cos wt+B\sin wt\) is \\ - $\dfrac{\omega}{2\pi}$ - $2 \pi \omega$ - $\dfrac{\omega}{2\pi}$ - $\dfrac{2\pi}{\omega}$ - \(A=(-4x-3y+az)\underline{i}+(bx+3y+5z)\underline{j}+(4
- Question 4, Exercise 1.3 @math-11-nbf:sol:unit01
- -i \end{align} \begin{align} \implies \omega & =\dfrac{1-i}{9+5i}\\ &=\dfrac{1-i}{9+5i}\times\dfrac{9-5i}{9-5i}\\ &=\dfrac{9-5-5i-9i}{81+25}\\ &=\dfrac{4-14i}{106}\\ &=\dfrac{2}{53}-\dfrac{7}{53}i\end{al
- Question 7, Exercise 10.2 @fsc-part1-kpk:sol:unit10
- tity ${{\cos }^{4}}\theta -{{\sin }^{4}}\theta =\dfrac{1}{\sec 2\theta }$. ====Solution==== \begin{align... uad \text{(By using double angle identity)}\\ &=\dfrac{1}{\sec 2\theta }=R.H.S.\end{align} =====Question 7(ii)===== Prove the identity $\tan \dfrac{\theta }{2}+co\operatorname{t}\dfrac{\theta }{2}=\dfrac{2}{\sin \theta }$. ====Solution==== \begin{alig
- Question 7, Exercise 10.2 @math-11-kpk:sol:unit10
- tity ${{\cos }^{4}}\theta -{{\sin }^{4}}\theta =\dfrac{1}{\sec 2\theta }$. ====Solution==== \begin{align... uad \text{(By using double angle identity)}\\ &=\dfrac{1}{\sec 2\theta }=R.H.S.\end{align} =====Question 7(ii)===== Prove the identity $\tan \dfrac{\theta }{2}+co\operatorname{t}\dfrac{\theta }{2}=\dfrac{2}{\sin \theta }$. ====Solution==== \begin{alig
- Exercise 1.2 (Solutions) @fsc-part1-ptb:sol:ch01
- tion 4(iv)** Simplify: $\displaystyle {{(-1)}^{-\frac{21}{2}}}$ **Solution** \begin{align} (-1)^{-\frac{21}{2}}&=\frac{1}{(-1)^\frac{21}{2}}=\frac{1}{[(-1)^\frac{1}{2}]^{21}}\\ &= \frac{1}{i^{21}}=\frac{1}{(i^2)^{10}\cdot
- Question 1 Exercise 5.3 @math-11-kpk:sol:unit05
- ==Question 1(i)==== Find the sum of the series $\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+\ldots$ to $n$ terms. ====Solution==== The general term of the series is: $$T_n=\dfrac{1}{n(n+1)}$$ Resolving $T_n$ into partial fractio
- 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: $\sin 2\theta=\dfrac{24}{25}$, $2\theta$ in QII. We have $$\cos 2\the... = - \sqrt{1-\sin^2 2\theta}\\ &=- \sqrt{1-\left(\frac{24}{25}\right)^2} \\ &=- \sqrt{\frac{49}{625}} = -\frac{7}{25} \end{align*} Also we have $$\sin\theta
- Question 7, Exercise 1.4 @math-11-nbf:sol:unit01
- lowing equation in Cartesian form: $\arg (z-1)=-\dfrac{\pi}{4}$ ** Solution. ** Suppose $z=x+iy$, as \begin{align*} &\arg (z-1)=-\dfrac{\pi}{4} \\ \implies & \arg(x+iy-1) = -\dfrac{\pi}{4} \\ \implies & \arg(x-1+iy) = -\dfrac{\pi}{4} \\ \implies & \tan^{-1}\left(\dfrac{y}{x-1}\right)
- 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 10.1 @fsc-part1-kpk:sol:unit10
- stan. =====Question 5(i)===== If $\tan \alpha =\dfrac{3}{4}$, $\sec \beta =\dfrac{13}{5}$ and neither the terminal side of the angle of measure $\alpha$ nor ... \right)$. ====Solution==== Given: $\tan\alpha =\dfrac{3}{4}$. As $\tan\alpha$ is +ive and terminal arm... \Rightarrow \quad \sec \alpha &=-\sqrt{1+{{\left(\frac{3}{4} \right)}^{2}}}=-\sqrt{1+\frac{9}{16}}=-\sqr