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- Question 7, Exercise 10.2 @fsc-part1-kpk:sol:unit10
- ====== Question 7, Exercise 10.2 ====== Solutions of Question 7 of Exercise 10.2 of Unit 10: Trigonometric I... \sin }^{4}}\theta =\dfrac{1}{\sec 2\theta }$. ====Solution==== \begin{align}L.H.S&={{\cos }^{4}}\theta -{{\s... }\dfrac{\theta }{2}=\dfrac{2}{\sin \theta }$. ====Solution==== \begin{align}L.H.S&=\tan \dfrac{\theta }{2}+c... eta }{1-\cos 2\theta }={{\cot }^{2}}\theta $. ====Solution==== \begin{align}L.H.S&=\dfrac{1+\cos 2\theta }{1
- Question 5, Exercise 1.3 @fsc-part1-kpk:sol:unit01
- ====== Question 5, Exercise 1.3 ====== Solutions of Question 5 of Exercise 1.3 of Unit 01: Complex Numbers. ... awar, Pakistan. =====Question 5(i)===== Find the solutions of the equation ${{z}^{2}}+z+3=0$\\ ====Solution==== ${{z}^{2}}+z+3=0$\\ According to the quadratic formu... 2}i\end{align} =====Question 5(ii)===== Find the solutions of the equation ${{z}^{2}}-1=z$.\\ ====Solution=
- Question 6, Exercise 1.3 @fsc-part1-kpk:sol:unit01
- ====== Question 6, Exercise 1.3 ====== Solutions of Question 6 of Exercise 1.3 of Unit 01: Complex Numbers. ... awar, Pakistan. =====Question 6(i)===== Find the solutions of the equation ${{z}^{4}}+{{z}^{2}}+1=0$\\ ====Solution==== \begin{align}{{z}^{4}}+{{z}^{2}}+1&=0\\ {{z}... }}}\end{align} =====Question 6(ii)===== Find the solutions of the equation ${{z}^{3}}=-8$\\ ====Solution===
- Question 7, Exercise 1.2 @fsc-part1-kpk:sol:unit01
- ====== Question 7, Exercise 1.2 ====== Solutions of Question 7 of Exercise 1.2 of Unit 01: Complex Numbers. ... eal and imaginary parts $\dfrac{2+3i}{5-2i}$. ====Solution==== \begin{align}&\dfrac{2+3i}{5-2i} \\ =&\dfrac... $\dfrac{{{\left( 1+2i \right)}^{2}}}{1-3i}$. ====Solution==== \begin{align}&\dfrac{(1+2i)^2}{1-3i}\\ =&\df... ts $\dfrac{1-i}{{{\left( 1+i \right)}^{2}}}$. ====Solution==== \begin{align}&\dfrac{1-i}{{{\left( 1+i \righ
- Question 1, Exercise 10.1 @fsc-part1-kpk:sol:unit10
- ====== Question 1, Exercise 10.1 ====== Solutions of Question 1 of Exercise 10.1 of Unit 10: Trigonometric ... }}+\cos {{37}^{\circ }}\sin {{22}^{\circ }}$ ==== Solution ==== As \begin{align} \sin (\alpha +\beta )=\si... }}+\sin {{83}^{\circ }}\sin {{53}^{\circ }}$ ====Solution==== As \begin{align}\cos (\alpha -\beta )&=\cos... }}-\sin {{19}^{\circ }}\sin {{5}^{\circ }}$ ==== Solution ==== As \begin{align}\cos (\alpha +\beta )&=\cos
- Question 2, Exercise 10.1 @fsc-part1-kpk:sol:unit10
- ====== Question 2, Exercise 10.1 ====== Solutions of Question 2 of Exercise 10.1 of Unit 10: Trigonometric ... === Evaluate exactly: $\sin \dfrac{\pi }{12}$ ===Solution=== We rewrite $\dfrac{\pi }{12}$ as $\dfrac{\pi }... ii)=== Evaluate exactly:$\tan {{75}^{\circ }}$ ==Solution== We rewrite ${{75}^{\circ }}$ as ${{45}^{\circ }... i)=== Evaluate exactly:$\tan {{105}^{\circ }}$ ==Solution== We rewrite ${{105}^{{}^\circ }}$ as ${{60}^{{}^
- Question 6, Exercise 10.2 @fsc-part1-kpk:sol:unit10
- ====== Question 6, Exercise 10.2 ====== Solutions of Question 6 of Exercise 10.2 of Unit 10: Trigonometric I... s to evaluate exactly $\cos {{15}^{\circ }}$. ====Solution==== Because ${{15}^{\circ }}=\dfrac{{{30}^{\circ ... to evaluate exactly $\tan {{67.5}^{\circ }}$. ====Solution==== Because ${{67.5}^{\circ }}=\dfrac{{{135}^{\ci... to evaluate exactly $sin{{112.5}^{\circ }}$. ====Solution==== Because ${{112.5}^{\circ }}=\dfrac{{{225}^{\c
- Unit 1: Complex Numbers (Solutions)
- ===== Unit 1: Complex Numbers (Solutions) ===== This is a first unit of the book Mathematics 11 published b... awar, Pakistan. On this page we have provided the solutions of the questions. After reading this unit the s... mber. <panel type="default" title="Exercise 1.1 (Solutions)"> * [[fsc-part1-kpk:sol:unit01:ex1-1-p1|Quest... panel> <panel type="default" title="Exercise 1.2 (Solutions)"> * [[fsc-part1-kpk:sol:unit01:ex1-2-p1|Quest
- Unit 10: Trigonometric Identities of Sum and Difference of Angles (Solutions)
- etric Identities of Sum and Difference of Angles (Solutions) ===== This is a tenth unit of the book Mathema... awar, Pakistan. On this page we have provided the solutions of the questions. After reading this unit the s... rdion><panel type="default" title="Exercise 10.1 (Solutions)"> * [[fsc-part1-kpk:sol:unit10:ex10-1-p1|Ques... anel> <panel type="default" title="Exercise 10.2 (Solutions)"> * [[fsc-part1-kpk:sol:unit10:ex10-2-p1|Qu
- Question 8, Exercise 1.2 @fsc-part1-kpk:sol:unit01
- ====== Question 8, Exercise 1.2 ====== Solutions of Question 8 of Exercise 1.2 of Unit 01: Complex Numbers. ... rline{z}=2\operatorname{Re}\left( z \right)$. ====Solution==== Assume $z=a+ib$, then $\overline{z}=a-ib$. \... line{z}=2i\operatorname{Im}\left( z \right)$. ====Solution==== Assume that $z=a+ib$, then $\overline{z}=a-i... ratorname{Im}\left( z \right) \right]}^{2}}$. ====Solution==== Suppose $z=a+ib$, then $\overline{z}=a-ib$.
- Question 2 & 3, Review Exercise 1 @fsc-part1-kpk:sol:unit01
- ====== Question 2 & 3, Review Exercise 1 ====== Solutions of Question 2 & 3 of Review Exercise 1 of Unit 01:... i}^{n+2}}+{{i}^{n+3}}=0$, $\forall n\in N$ \\ ====Solution==== \begin{align}{{i}^{n}}+{{i}^{n+1}}+{{i}^{n+2... left( 5+7i \right)$ in the form of $x+iy$.\\ ====Solution==== $\left( 1+3i \right)+\left( 5+7i \right)=1+5... left( 5+7i \right)$ in the form of $x+iy$.\\ ====Solution==== \begin{align}\left( 1+3i \right)-\left( 5+7i
- Question 1, Exercise 1.1 @fsc-part1-kpk:sol:unit01
- ====== Question 1, Exercise 1.1 ====== Solutions of Question 1 of Exercise 1.1 of Unit 01: Complex Numbers. ... on 1(i)===== Simplify ${{i}^{9}}+{{i}^{19}}$. ====Solution==== \begin{align}{{i}^{9}}+{{i}^{19}}&=i\cdot{{i}... )===== Simplify ${{\left( -i \right)}^{23}}$. ====Solution==== \begin{align}{{\left( -i \right)}^{23}}&={{\l... lify ${{\left( -1 \right)}^{\frac{-23}{2}}}$. ====Solution==== \begin{align}{{\left( -1 \right)}^{\frac{-23}
- Question 2 & 3, Exercise 1.1 @fsc-part1-kpk:sol:unit01
- ====== Question 2 & 3, Exercise 1.1 ====== Solutions of Question 2 & 3 of Exercise 1.1 of Unit 01: Complex N... 107}}+{{i}^{112}}+{{i}^{122}}+{{i}^{153}}=0$. ====Solution==== \begin{align}L.H.S.&={{i}^{107}}+{{i}^{112}}+... $3\left( 1+2i \right),-2\left( 1-3i \right)$. ====Solution==== \begin{align}& 3\left( 1+2i \right)+-2\left( ... 2}-\dfrac{2}{3}i,\dfrac{1}{4}-\dfrac{1}{3}i$. ====Solution==== \begin{align}&\left( \dfrac{1}{2}-\dfrac{2}{3
- Question 6, Exercise 1.1 @fsc-part1-kpk:sol:unit01
- ====== Question 6, Exercise 1.1 ====== Solutions of Question 6 of Exercise 1.1 of Unit 01: Complex Numbers. ... i}$ and write the answer in the form $a+ib$. ====Solution==== \begin{align}\dfrac{4+i}{3+5i}&=\dfrac{4+i}{3... i}$ and write the answer in the form $a+ib$. ====Solution==== \begin{align}\dfrac{1}{-8+i}&=\dfrac{1}{-8+i}... i}$ and write the answer in the form $a+ib$. ====Solution==== \begin{align}\dfrac{1}{7-3i}&=\dfrac{1}{7-3i}
- Question 2, Exercise 1.3 @fsc-part1-kpk:sol:unit01
- ====== Question 2, Exercise 1.3 ====== Solutions of Question 2 of Exercise 1.3 of Unit 01: Complex Numbers. ... actors. $$P\left( z \right)={{z}^{3}}+6z+20$$ ====Solution==== Given: $$p\left( z \right)={{z}^{3}}+6z+20$... $P(z)$ into linear factors. $$P(z)=3z^2+7.$$ ====Solution==== \begin{align} P(z)&=3z^2+7\\ &=\left(\sqrt{3}... r factors. $$P\left( z \right)={{z}^{2}}+4$$ ====Solution==== \begin{align}P(z)&={{z}^{2}}+4\\ &={{\left(