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- Question 1, Exercise 1.3 @fsc-part1-kpk:sol:unit01
- plex coefficient. \begin{align}&z-4w=3i\\ &2z+3w=11-5i\end{align} ====Solution==== Given that \begin{align}z-4w&=3i …(i)\\ 2z+3w&=11-5i …(ii)\end{align} Multiply $2$ by (i), we get\\... s_{-}3w&=\mathop-\limits_{+}5i&\mathop+\limits_{-}11 \\ \hline 0&-11w&=11i &-11\\ \end{array} \] \begin{align}-11w&=11i-11\\ \implies w&=\dfrac{11-11i}{
- Question 11, Exercise 1.1 @fsc-part1-kpk:sol:unit01
- ====== Question 11, Exercise 1.1 ====== Solutions of Question 11 of Exercise 1.1 of Unit 01: Complex Numbers. This is un... (KPTB or KPTBB) Peshawar, Pakistan. =====Question 11(i)===== Let ${{z}_{1}}=2-i$, ${{z}_{2}}=-2+i$. ... i}{2-i}\\ &=\dfrac{-6+4+8i+3i}{4+1}\\ &=\dfrac{-2+11i}{5}\\ &=\dfrac{-2}{5}+\dfrac{11i}{5} \end{align}
- Question 1, Exercise 1.1 @fsc-part1-kpk:sol:unit01
- 1 \right)\cdot i\cdot{{\left( {{i}^{2}} \right)}^{11}}\\ &=-i\cdot{{\left( {{i}^{2}} \right)}^{11}}\\ &=-i\cdot{{\left( -1 \right)}^{11}}\\ &=-i\cdot\left( -1 \right)\\ &=i\end{align} =====Question ... {-23}}\\ &=\frac{1}{i{{\left( {{i}^{2}} \right)}^{11}}}\\ &=\frac{1}{i{{\left( -1 \right)}^{11}}}\\ &=
- Question 5, Exercise 1.1 @fsc-part1-kpk:sol:unit01
- uestion 5(i)===== Multiply the complex number $8i+11,-7+5i$. ====Solution==== \begin{align}&(8i+11)\times (-7+5i)\\ &=\left( 11+8i \right)\times \left( -7+5i \right)\\ &=\left( -77+40{{i}^{2}} \right)... =\left( -77-40 \right)+\left( 55-56 \right)i\\ &=-117-i\end{align} =====Question 5(ii)===== Multiply
- Question 2, Exercise 10.2 @fsc-part1-kpk:sol:unit10
- [4px,border:2px solid black]{\cos 2\theta=-\dfrac{119}{169}.}$$ =====Question 2(iii)===== If $\sin ... ^{2}}\\ &=\dfrac{144}{169}-\dfrac{25}{169}=\dfrac{119}{169}.\end{align} Now \begin{align}\tan 2\theta... \cos 2\theta }\\ &=\dfrac{\frac{-120}{169}}{\frac{119}{169}}\end{align} $$\implies \bbox[4px,border:2px solid black]{\tan 2\theta=-\dfrac{120}{119}}$$ <text align="left"><btn type="primary">[[fs
- Question 6, Exercise 10.2 @fsc-part1-kpk:sol:unit10
- e half angle identities to evaluate exactly $sin{{112.5}^{\circ }}$. ====Solution==== Because ${{112.5}^{\circ }}=\dfrac{{{225}^{\circ }}}{2}$, the $\dfra... dfrac{{{225}^{\circ }}}{2}$, so we can find $sin{{112.5}^{\circ }}$by using half angle identity as, \begin{align} sin{{112.5}^{\circ }}&=\sin \dfrac{{{225}^{\circ }}}{2}=\
- Question 2 & 3, Exercise 1.1 @fsc-part1-kpk:sol:unit01
- ====Question 2===== Prove that ${{i}^{107}}+{{i}^{112}}+{{i}^{122}}+{{i}^{153}}=0$. ====Solution==== \begin{align}L.H.S.&={{i}^{107}}+{{i}^{112}}+{{i}^{122}}+{{i}^{153}}\\ &=i\cdot i^{106}+i^{112}+i^{122}+i\cdot i^{152}\\ &=i.{{\left( {{i}^{2}}
- Question 5, Exercise 1.3 @fsc-part1-kpk:sol:unit01
- rac{-1\pm \sqrt{1-12}}{2}\\ z&=\dfrac{-1\pm \sqrt{11}}{2}i\\ z&=-\dfrac{1}{2}+\dfrac{\sqrt{11}}{2}i,-\dfrac{1}{2}-\dfrac{\sqrt{11}}{2}i\end{align} =====Question 5(ii)===== Find the solutions of
- Unit 1: Complex Numbers (Solutions)
- === This is a first unit of the book Mathematics 11 published by Khyber Pakhtunkhwa Textbook Board, P... * [[fsc-part1-kpk:sol:unit01:ex1-1-p9|Question 11]] </panel> <panel type="default" title="Exercise
- Unit 10: Trigonometric Identities of Sum and Difference of Angles (Solutions)
- === This is a tenth unit of the book Mathematics 11 published by Khyber Pakhtunkhwa Textbook Board, P... * [[fsc-part1-kpk:sol:unit10:ex10-1-p10|Question 11-12]] * [[fsc-part1-kpk:sol:unit10:ex10-1-p11|Qu
- Question11 and 12, Exercise 10.1 @fsc-part1-kpk:sol:unit10
- d 12, Exercise 10.1 ====== Solutions of Question 11 and 12 of Exercise 10.1 of Unit 10: Trigonometric... (KPTB or KPTBB) Peshawar, Pakistan. =====Question 11===== If $\alpha$, $\beta$, $\gamma$ are the angle
- Question 8, Exercise 1.1 @fsc-part1-kpk:sol:unit01
- -7i}{4-7i}\\ &=\dfrac{\left( 64-14 \right)-\left( 112+8 \right)i}{16+49}\\ &=\dfrac{50-120i}{65}\\ &=\
- Question 9 & 10, Exercise 1.1 @fsc-part1-kpk:sol:unit01
- t align="right"><btn type="success">[[fsc-part1-kpk:sol:unit01:ex1-1-p9|Question 11 >]]</btn></text>
- Question, Exercise 10.1 @fsc-part1-kpk:sol:unit10
- - \dfrac{5}{12}}{1+ \dfrac{20}{36}}=\dfrac{\dfrac{11}{12}}{ \dfrac{56}{36}}\\ \Rightarrow \quad \tan \
- Question 9 and 10, Exercise 10.1 @fsc-part1-kpk:sol:unit10
- gn="right"><btn type="success">[[fsc-part1-kpk:sol:unit10:ex10-1-p10|Question 11,12 >]]</btn></text>