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- Question 5, Exercise 10.1 @fsc-part1-kpk:sol:unit10
- ad \sec \alpha &=\pm \sqrt{1+{{\tan}^{2}}\alpha}.\end{align} Since terminal arm of $\alpha$ is in the ... ,}\\ \Rightarrow \quad \sec \alpha&=-\frac{5}{4} \end{align} Now $$\cos\alpha =\frac{1}{\sec \alpha }=... t)\\ \Rightarrow \quad \sin\alpha &= \frac{3}{5}\end{align} Given: $\sec \beta =\dfrac{13}{5}.$ As... frac{13}{5}}\\ \implies \cos\beta &=\dfrac{5}{13}\end{align} As $\cos\beta $ is +ive and terminal arm
- Question 1, Exercise 1.3 @fsc-part1-kpk:sol:unit01
- oefficient. \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\\ \begin{align}2z-8w&=6i …(iii)\end{align} Subtract (iii) from (ii), we get\\ \[\begi... thop+\limits_{-}11 \\ \hline 0&-11w&=11i &-11\\ \end{array} \] \begin{align}-11w&=11i-11\\ \implies w
- Question 13, Exercise 10.1 @fsc-part1-kpk:sol:unit10
- ht) = 16+9\\ \implies &r^2 = 25\\ \implies &r=5 \end{align} Also \begin{align}r\cos\varphi = 4 \implie... cos\varphi = 4 \implies \cos\varphi =\dfrac{4}{5}\end{align} and \begin{align}r\sin\varphi = 3 \implies... in\varphi = 3 \implies \sin\varphi =\dfrac{3}{5}.\end{align} Thus, from (1), we have \begin{align}&4\si... 3}{5}, \cos\varphi=\dfrac{4}{5} \text{ and } r=5.\end{align} =====Question 13(ii)===== Express each of
- Question 7, Exercise 10.2 @fsc-part1-kpk:sol:unit10
- le identity)}\\ &=\dfrac{1}{\sec 2\theta }=R.H.S.\end{align} =====Question 7(ii)===== Prove the identi... quad (By \,using\, double\, angle\, identity)\end{align} =====Question 7(iii)===== Prove the ident... {2si{{n}^{2}}\theta }={{\cot }^{2}}\theta =R.H.S.\end{align} =====Question 7(iv)===== Prove the identi... }{2\sin \theta \cos \theta }=\tan \theta =R.H.S.\end{align} =====Question 7(v)===== Prove the identi
- Question 3, Exercise 10.1 @fsc-part1-kpk:sol:unit10
- 25}}\\ \Rightarrow \,\,\,\,\,\cos u&=\dfrac{4}{5}\end{align} Also $\cos v=\sqrt{1-{{\sin }^{2}}v}$ As $... {9}{25}}\\ \Rightarrow \,\,\,\cos v &=\frac{3}{5}\end{align} Now \begin{align}\cos \left( u+v \right)... 2}{25}-\frac{12}{25}\\ \cos \left( u+v \right)&=0\end{align} =====Question 3(ii)===== If $\sin u=\df... \dfrac{4}{5},\,\,\,\,\,0\le v\le \dfrac{\pi }{2}.\end{align} We know $\cos u=\pm \sqrt{1-{{\sin }^{2}}u
- Question 1, Exercise 10.1 @fsc-part1-kpk:sol:unit10
- =\sin \alpha \cos \beta +\cos \alpha \sin \beta, \end{align} Therefore \begin{align} \sin {{37}^{\cir... \left( 37+22 \right) \\ & =\sin {{59}^{\circ }}. \end{align} ===== Question 1(ii)===== Write as a tr... &=\cos \alpha \cos \beta +\sin \alpha \sin \beta,\end{align} Therefore \begin{align}\cos {{83}^{\circ ... cos \left( 83-53 \right)\\ &=\cos {{30}^{\circ }}\end{align}. ===== Question 1(iii)===== Write as a
- Question 5, Exercise 1.2 @fsc-part1-kpk:sol:unit01
- i$. Now \begin{align}z_1+z_2&=2+4i+1-3i\\ &=3+i \end{align} Now \begin{align}\overline{z_1+z_2}=3-i \ldots (1)\end{align} and \begin{align} \overline{z_1}+\overline{z_2}&=2-4i+1+3i\\ &=3-i \ldots (2)\end{align} From (1) and (2), we have $$\overline{z_1... lign}z_1 z_2 &=(2+3i)(2-3i)\\ &=2^2-(3i)^2\\ &=13\end{align} Now \begin{align}\overline{z_1 z_2}=13 \ld
- Question 6, Exercise 1.3 @fsc-part1-kpk:sol:unit01
- z}^{2}}+\dfrac{1}{2} \right)}^{2}}&=-\dfrac{3}{4}\end{align}\\ Take square root on both sides.\\ \begin... \pm \dfrac{\sqrt{3}}{2}i \right)}^{\dfrac{1}{2}}}\end{align} =====Question 6(ii)===== Find the solutio... {{a}^{2}}-ab+{{b}^{2}} \right)\\ z+2&=0\\ z&=-2\end{align} Now\\ $\left( {{z}^{2}}-2z+4 \right)=0$\... z&=\dfrac{2\pm 2\sqrt{3}i}{2}\\ z&=1\pm \sqrt{3}i\end{align} Then\\ $$z=-2,1\pm \sqrt{3}i$$ =====Quest
- Question, Exercise 10.1 @fsc-part1-kpk:sol:unit10
- }\\ \Rightarrow \quad \cos \alpha&=-\dfrac{3}{5}.\end{align} Also $$\sin \beta=\pm\sqrt{1-\cos^2\beta}.... }}\\ \Rightarrow \quad \sin \beta &=\frac{5}{13}.\end{align} Now \begin{align}\sin (\alpha-\beta )&=\s... \cos \left( \alpha+\beta \right)&=\frac{33}{65}.\end{align} =====Question 4(ii)===== If $\sin \alpha... }\\ \Rightarrow \quad \cos \alpha&=-\dfrac{3}{5}.\end{align} Also $$\sin \beta=\pm\sqrt{1-\cos^2\beta}.
- Question 2, Exercise 1.2 @fsc-part1-kpk:sol:unit01
- &=\left( -1+i \right)+\left( 3-2i \right)\\ &=2-i\end{align} So \begin{align} \left( {{z}_{1}}+{{z}_{2}... i \right)+\left( 2-2i \right)\\ &=4-3i \ldots (1)\end{align} Now \begin{align} {{z}_{2}}+{{z}_{3}}&=\left( 3-2i \right)+\left( 2-2i \right)\\ &=5-4i\end{align} So \begin{align} {{z}_{1}}+\left({{z}_{2}}... \right)+\left( 5-4i \right)\\ &=4-3i \ldots (2) \end{align} From (1) and (2), we have the required res
- Question 2, Exercise 1.3 @fsc-part1-kpk:sol:unit01
- {align} P(-2)&=(-2)^3+6(-2)+20\\ &=-8-12+20\\ &=0\end{align} Thus $z+2$ is a factor of ${{z}^{3}}+6z+2... & -2 & 4 & -20 \\ \hline & 1 & -2 & 10 & 0 \\ \end{array}$$ This gives \begin{align} P(z)&=(z+2)(z^2... t[(z-1)^2-(3i)^2\right]\\ &=(z+2)(z-1+3i)(z-1-3i)\end{align} =====Question 2(ii)===== Factorize the po... qrt{7}i \right)\left( \sqrt{3}z-\sqrt{7}i \right)\end{align} =====Question 2(iii)===== Factorize the
- Question 2, Exercise 10.2 @fsc-part1-kpk:sol:unit10
- (\frac{5}{13}\right)}\\ &=-\sqrt{\frac{144}{169}}\end{align} $$\implies \cos\theta = -\dfrac{12}{13}$$ ... dfrac{5}{13} \right)\left( \dfrac{12}{13} \right)\end{align} $$\implies \bbox[4px,border:2px solid blac... (\frac{5}{13}\right)}\\ &=-\sqrt{\frac{144}{169}}\end{align} $$\implies \cos\theta = -\dfrac{12}{13}$$ ... right)}^{2}}\\ &=\dfrac{144}{169}-\dfrac{25}{169}\end{align} $$\implies \bbox[4px,border:2px solid blac
- Question 8, Exercise 1.2 @fsc-part1-kpk:sol:unit01
- +\overline{z}&=2\operatorname{Re}\left( z \right)\end{align} =====Question 8(ii)===== Show that $z-\o... {z}&=2bi\\ z-\overline{z}&=2i\operatorname{Im}(z)\end{align} =====Question 8(iii)===== Show that $z\o... mplies z\overline{z}&={{a}^{2}}+b^2. \ldots (1) \end{align} Now \begin{align} {{\left[\operatorname{Re... ) \right]}^{2}}&={{a}^{2}}+{{b}^{2}}. \ldots (2) \end{align} Using (1) and (2), we get $$z\overline{z}
- Question 2, Exercise 10.1 @fsc-part1-kpk:sol:unit10
- \beta )=\sin \alpha \cos \beta -\cos \alpha \sin.\end{align} \begin{align} \Rightarrow \quad \sin \lef... c{\sqrt{2}}{4} \\ &=\frac{\sqrt{6}-\sqrt{2}}{4}. \end{align} ===Question 2(ii)=== Evaluate exactly:$... sqrt{3} \right)}{2}\\ &=2+\left( \sqrt{3} \right)\end{align} ===Question 2(iii)=== Evaluate exactly:$... 3}{-2}\\ &=\dfrac{4+\sqrt{3}}{-2}\\ &=-2-\sqrt{3}\end{align} ===Question 2(iv)=== Evaluate exactly:$\t
- Question11 and 12, Exercise 10.1 @fsc-part1-kpk:sol:unit10
- a}{2}+\dfrac{\beta}{2}=90^\circ-\dfrac{\gamma}{2}\end{align} Now \begin{align}&\tan\left( \dfrac{\alpha... 90-\tfrac{\gamma}{2}\right)=\cot\tfrac{\gamma}{2}\end{align} \begin{align} \implies & \frac{\tan\dfrac{... pha}{2}\cot\dfrac{\beta}{2}\cot\dfrac{\gamma}{2}.\end{align} =====Question 12===== If $\alpha +\beta... irc\\ \implies &\alpha +\beta=180^\circ-\gamma. \end{align} Now \begin{align} & \tan \left( \alpha +\