Search
You can find the results of your search below.
Fulltext results:
- Question 4, Exercise 1.3 @math-11-nbf:sol:unit01
- 1-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{ali
- Question 7, Exercise 10.2 @fsc-part1-kpk:sol:unit10
- ntity ${{\cos }^{4}}\theta -{{\sin }^{4}}\theta =\dfrac{1}{\sec 2\theta }$. ====Solution==== \begin{align... quad \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{align
- Question 7, Exercise 10.2 @math-11-kpk:sol:unit10
- ntity ${{\cos }^{4}}\theta -{{\sin }^{4}}\theta =\dfrac{1}{\sec 2\theta }$. ====Solution==== \begin{align... quad \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{align
- 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 2, Exercise 2.3 @math-11-kpk:sol:unit02
- eft| \begin{matrix} -3 & 4 & -5 \\ 1 & -1 & 2 \\ \dfrac{5}{49} & -\dfrac{6}{49} & \dfrac{8}{49}\end{matrix} \right. \right]\text{by}-\dfrac{1}{49}R_3\\ \underset{\sim}{R}&\left[\begin{matrix} 1
- Question 4 Review Exercise @math-11-kpk:sol:unit05
- Pakistan. =====Question 4===== Sum the series: $\dfrac{1}{1.4 .7}+\dfrac{1}{4.7 .10}+\dfrac{1}{7.10 .13}+\ldots$ ====Solution==== In the denominator Each term is the product of ... $ Thus the general term of the series is: $$a_n=\dfrac{1}{(3 n-2)(3 n+1)(3 n+4)}$$ Resolving into partia
- PPSC Paper 2015 (Lecturer in Mathematics) @ppsc
- 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}+(4x... semicircular wire of radius \(a\) is \\ - \(\dfrac{2a}{\pi} \) - \(\dfrac{4a}{\pi} \) - \(\d
- Exercise 1.2 (Solutions) @fsc-part1-ptb:sol:ch01
- 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(
- Question 2, Exercise 10.1 @fsc-part1-kpk:sol:unit10
- ====Question 2(i)==== Evaluate exactly: $\sin \dfrac{\pi }{12}$ ===Solution=== We rewrite $\dfrac{\pi }{12}$ as $\dfrac{\pi }{3}-\dfrac{\pi }{4}$ and using the identity: \begin{align}\sin (\alpha -\beta )=\sin \alpha \cos \b
- Question 2, Exercise 10.1 @math-11-kpk:sol:unit10
- ====Question 2(i)==== Evaluate exactly: $\sin \dfrac{\pi }{12}$ ===Solution=== We rewrite $\dfrac{\pi }{12}$ as $\dfrac{\pi }{3}-\dfrac{\pi }{4}$ and using the identity: \begin{align}\sin (\alpha -\beta )=\sin \alpha \cos \b
- Question 5, Exercise 10.3 @fsc-part1-kpk:sol:unit10
- \circ }}\cos {{60}^{\circ }}\cos {{80}^{\circ }}=\dfrac{1}{16}.$$ ====Solution==== We know that\\ $2\cos ... &=\cos {{20}^{\circ }}\cos {{40}^{\circ }}\left( \dfrac{1}{2} \right)\cos {{80}^{\circ }}\\ &=\dfrac{1}{2}\cos {{80}^{\circ }}\cos {{40}^{\circ }}\cos {{20}^{\circ }}\\ &=\dfrac{1}{4}\left( 2\,\cos {{80}^{\circ }}\cos {{40}^{\c
- Question 5, Exercise 10.3 @fsc-part1-kpk:sol:unit10
- \circ }}\cos {{60}^{\circ }}\cos {{80}^{\circ }}=\dfrac{1}{16}$. ====Solution==== We know that\\ $2\cos \... &=\cos {{20}^{\circ }}\cos {{40}^{\circ }}\left( \dfrac{1}{2} \right)\cos {{80}^{\circ }}\\ &=\dfrac{1}{2}\cos {{80}^{\circ }}\cos {{40}^{\circ }}\cos {{20}^{\circ }}\\ &=\dfrac{1}{4}\left( 2\,\cos {{80}^{\circ }}\cos {{40}^{\c
- Question 5, Exercise 10.3 @math-11-kpk:sol:unit10
- \circ }}\cos {{60}^{\circ }}\cos {{80}^{\circ }}=\dfrac{1}{16}.$$ ====Solution==== We know that\\ $2\cos ... &=\cos {{20}^{\circ }}\cos {{40}^{\circ }}\left( \dfrac{1}{2} \right)\cos {{80}^{\circ }}\\ &=\dfrac{1}{2}\cos {{80}^{\circ }}\cos {{40}^{\circ }}\cos {{20}^{\circ }}\\ &=\dfrac{1}{4}\left( 2\,\cos {{80}^{\circ }}\cos {{40}^{\c
- Question 5, Exercise 10.3 @math-11-kpk:sol:unit10
- \circ }}\cos {{60}^{\circ }}\cos {{80}^{\circ }}=\dfrac{1}{16}$. ====Solution==== We know that\\ $2\cos \... &=\cos {{20}^{\circ }}\cos {{40}^{\circ }}\left( \dfrac{1}{2} \right)\cos {{80}^{\circ }}\\ &=\dfrac{1}{2}\cos {{80}^{\circ }}\cos {{40}^{\circ }}\cos {{20}^{\circ }}\\ &=\dfrac{1}{4}\left( 2\,\cos {{80}^{\circ }}\cos {{40}^{\c
- Question 6, Exercise 10.2 @fsc-part1-kpk:sol:unit10
- c }}$. ====Solution==== Because ${{15}^{\circ }}=\dfrac{{{30}^{\circ }}}{2}$, and $\dfrac{\theta }{2}=\dfrac{{{30}^{\circ }}}{2}$, we can find $\cos {{15}^{\circ }}$by using half angle identity as, \begin{align}\cos {{15}^{\circ }}&=\cos \dfrac{{{30}^{\circ }}}{2}=\sqrt{\dfrac{1+\cos {{30}^{\c