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- Question 7, Exercise 10.2 @fsc-part1-kpk:sol:unit10
- n} <text align="left"><btn type="primary">[[fsc-part1-kpk:sol:unit10:ex10-2-p5|< Question 6]]</btn></text> <text align="right"><btn type="success">[[fsc-part1-kpk:sol:unit10:ex10-2-p7|Question 8, 9 >]]</btn></text>
- Question 7, Exercise 1.2 @fsc-part1-kpk:sol:unit01
- estion 7(i)===== Separate into real and imaginary parts $\dfrac{2+3i}{5-2i}$. ====Solution==== \begin{align}&\dfrac{2+3i}{... =&\dfrac{4}{29}+\dfrac{19}{29}i \end{align} Real part $=\dfrac{4}{29}$\\ Imaginary part $=\dfrac{19}{29}$ =====Question 7(ii)===== Separate into real and imaginary parts $\dfrac{{{\left( 1+2i \right)}^{2}}}{1-3i}$. ====Solution==== \begin{align}&\dfrac
- Question 6, Exercise 1.2 @fsc-part1-kpk:sol:unit01
- ==== <text align="left"><btn type="primary">[[fsc-part1-kpk:sol:unit01:ex1-2-p4|< Question 5]]</btn></text> <text align="right"><btn type="success">[[fsc-part1-kpk:sol:unit01:ex1-2-p6|Question 7 >]]</btn></text>
- Question 3 & 4, Exercise 1.2 @fsc-part1-kpk:sol:unit01
- ==== <text align="left"><btn type="primary">[[fsc-part1-kpk:sol:unit01:ex1-2-p2|< Question 2]]</btn></text> <text align="right"><btn type="success">[[fsc-part1-kpk:sol:unit01:ex1-2-p4|Question 5 >]]</btn></text>
- Question 5, Exercise 1.2 @fsc-part1-kpk:sol:unit01
- ==== <text align="left"><btn type="primary">[[fsc-part1-kpk:sol:unit01:ex1-2-p3|< Question 3 & 4]]</btn></text> <text align="right"><btn type="success">[[fsc-part1-kpk:sol:unit01:ex1-2-p5|Question 6 >]]</btn></text>
- Question 8 and 9, Exercise 10.2 @fsc-part1-kpk:sol:unit10
- -1}=R.H.S.\end{align} <text align="left"><btn type="primary">[[fsc-part1-kpk:sol:unit10:ex10-2-p6|< Question 7]]</btn></text>
- Question 6, Exercise 10.2 @fsc-part1-kpk:sol:unit10
- <text align="left"><btn type="primary">[[fsc-part1-kpk:sol:unit10:ex10-2-p4|< Question 4, 5]]</btn></text> <text align="right"><btn type="success">[[fsc-part1-kpk:sol:unit10:ex10-2-p6|Question 7 >]]</btn></text>
- 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 Identities... (KPTB or KPTBB) Peshawar, Pakistan. ====Question 2(i)==== Evaluate exactly: $\sin \dfrac{\pi }{12}$ ===Solution=== We rewrite $\dfrac{\pi }{12}$ as $\
- Question 13, Exercise 10.1 @fsc-part1-kpk:sol:unit10
- } \text{ and } r=\sqrt{2}.\end{align} <text align="left"><btn type="primary">[[fsc-part1-kpk:sol:unit10:ex10-1-p10|< Question 11,12]]</btn></text>
- Question 6, Exercise 1.3 @fsc-part1-kpk:sol:unit01
- nd the solutions of the equation ${{z}^{4}}+{{z}^{2}}+1=0$\\ ====Solution==== \begin{align}{{z}^{4}}+{{z}^{2}}+1&=0\\ {{z}^{4}}+2\left( \dfrac{1}{2} \right){{z}^{2}}+\dfrac{1}{4}-\dfrac{1}{4}+1&=0\\ {{\left( {{z}^{2}}+\dfrac{1}{2}
- Question 5, Exercise 10.3 @fsc-part1-kpk:sol:unit10
- stan. =====Question 5(i)===== Prove that $$\cos {{20}^{\circ }}\cos {{40}^{\circ }}\cos {{60}^{\circ ... \dfrac{1}{16}.$$ ====Solution==== We know that\\ $2\cos \alpha \cos \beta =\cos \left( \alpha +\beta ... ha -\beta \right)$\\ \begin{align}L.H.S.&=\cos {{20}^{\circ }}\cos {{40}^{\circ }}\cos {{60}^{\circ }}\cos {{80}^{\circ }}\\ &=\cos {{20}^{\circ }}\cos {{40}^{\circ }}\left( \dfrac{1}{2
- 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. This is un... KPTB or KPTBB) Peshawar, Pakistan. =====Question 2(i)===== Factorize the polynomial $P(z)$ into linear factors. $$P\left( z \right)={{z}^{3}}+6z+20$$ ====Solution==== Given: $$p\left( z \right)=
- Question 2, Exercise 10.3 @fsc-part1-kpk:sol:unit10
- ====== Question 2, Exercise 10.3 ====== Solutions of Question 2 of Exercise 10.3 of Unit 10: Trigonometric Identities ... (KPTB or KPTBB) Peshawar, Pakistan. =====Question 2(i)===== Convert the sum or difference as product:... We have an identity: $$\sin \alpha +\sin \beta =2\sin \left( \dfrac{\alpha +\beta }{2} \right)\cos
- Question 5, Exercise 10.3 @fsc-part1-kpk:sol:unit10
- istan. =====Question 5(i)===== Prove that $\cos {{20}^{\circ }}\cos {{40}^{\circ }}\cos {{60}^{\circ ... =\dfrac{1}{16}$. ====Solution==== We know that\\ $2\cos \alpha \cos \beta =\cos \left( \alpha +\beta ... ha -\beta \right)$\\ \begin{align}L.H.S.&=\cos {{20}^{\circ }}\cos {{40}^{\circ }}\cos {{60}^{\circ }}\cos {{80}^{\circ }}\\ &=\cos {{20}^{\circ }}\cos {{40}^{\circ }}\left( \dfrac{1}{2
- Question 3, Exercise 10.1 @fsc-part1-kpk:sol:unit10
- n} <text align="left"><btn type="primary">[[fsc-part1-kpk:sol:unit10:ex10-1-p2|< Question 2]]</btn></text> <text align="right"><btn type="success">[[fsc-part1-kpk:sol:unit10:ex10-1-p4|Question 4 >]]</btn></t