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- Question 5, Exercise 10.1 @fsc-part1-kpk:sol:unit10
- quadrant, therefor it lies in 3rd quadrant. Now \begin{align}{{\sec}^{2}}\alpha &=1+{{\tan}^{2}}\alpha\\... is in the 3rd quadrant, value of $\sec$ is –ive \begin{align}\sec \alpha &=-\sqrt{1+{{\tan}^{2}}\alpha }... \Rightarrow \quad \cos\alpha=-\dfrac{4}{5}$$ Now \begin{align}\frac{\sin \alpha }{\cos \alpha }&=\tan \al... ign} Given: $\sec \beta =\dfrac{13}{5}.$ As \begin{align}\cos\beta &= \dfrac{1}{\sec\beta} = \dfrac{
- Question 1, Exercise 1.3 @fsc-part1-kpk:sol:unit01
- aneous linear equation with complex 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\\ \begin{align}2z-8w&=6i …(iii)\end{align} Subtract (iii) from (ii), we get\\ \[\begin{array}{cccc} 2z&-8w&=6i \\ \mathop+\limits_{-}
- Question 13, Exercise 10.1 @fsc-part1-kpk:sol:unit10
- phi$ and $3=r\sin\varphi$. Squaring and adding \begin{align} &r^2\cos^2 \varphi+r^2\sin^2\varphi = 4^2+... lies &r^2 = 25\\ \implies &r=5 \end{align} Also \begin{align}r\cos\varphi = 4 \implies 5\cos\varphi = 4 ... implies \cos\varphi =\dfrac{4}{5}\end{align} and \begin{align}r\sin\varphi = 3 \implies 5\sin\varphi = 3 ... \dfrac{3}{5}.\end{align} Thus, from (1), we have \begin{align}&4\sin\theta +3\cos\theta \\ &=5 \left(\dfr
- Question 7, Exercise 10.2 @fsc-part1-kpk:sol:unit10
- eta =\dfrac{1}{\sec 2\theta }$. ====Solution==== \begin{align}L.H.S&={{\cos }^{4}}\theta -{{\sin }^{4}}\t... }{2}=\dfrac{2}{\sin \theta }$. ====Solution==== \begin{align}L.H.S&=\tan \dfrac{\theta }{2}+co\operatorn... \theta }={{\cot }^{2}}\theta $. ====Solution==== \begin{align}L.H.S&=\dfrac{1+\cos 2\theta }{1+\cos 2\the... eta -\cot 2\theta =tan\theta $. ====Solution==== \begin{align}L.H.S.&=\cos ec2\theta -\cot 2\theta \\ &=
- Question 3, Exercise 10.1 @fsc-part1-kpk:sol:unit10
- uadrant and $\cos $ is $+ve$ in first quadrant . \begin{align}\Rightarrow \,\,\,\,\,\cos u&=\sqrt{1-{{\si... quadrant and $\cos $ is $+ve$ in first quadrant. \begin{align}\Rightarrow \,\,\,\,\,\cos v&=\sqrt{1-{{\si... rrow \,\,\,\cos v &=\frac{3}{5}\end{align} Now \begin{align}\cos \left( u+v \right)&=\cos u\cos v-\sin ... \tan \left( u-v \right)$ ====Solution==== Given \begin{align}\sin u&=\dfrac{3}{5},\,\,\,\, 0\le u\le
- Question 1, Exercise 10.1 @fsc-part1-kpk:sol:unit10
- }}\sin {{22}^{\circ }}$ ==== Solution ==== As \begin{align} \sin (\alpha +\beta )=\sin \alpha \cos \be... +\cos \alpha \sin \beta, \end{align} Therefore \begin{align} \sin {{37}^{\circ }}\cos {{22}^{\circ }}+\... rc }}\sin {{53}^{\circ }}$ ====Solution==== As \begin{align}\cos (\alpha -\beta )&=\cos \alpha \cos \be... a +\sin \alpha \sin \beta,\end{align} Therefore \begin{align}\cos {{83}^{\circ }}\cos {{53}^{\circ }}+\s
- Question 5, Exercise 1.2 @fsc-part1-kpk:sol:unit01
- 1}}}=2-4i$ and $\overline{{{z}_{2}}}=1+3i$. 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\\... z}_{1}}}=2-3i$ and $\overline{{{z}_{2}}}=2+3i$. \begin{align}z_1 z_2 &=(2+3i)(2-3i)\\ &=2^2-(3i)^2\\ &=1
- Question 6, Exercise 1.3 @fsc-part1-kpk:sol:unit01
- on ${{z}^{4}}+{{z}^{2}}+1=0$\\ ====Solution==== \begin{align}{{z}^{4}}+{{z}^{2}}+1&=0\\ {{z}^{4}}+2\left... }\end{align}\\ Take square root on both sides.\\ \begin{align}\left( {{z}^{2}}+\dfrac{1}{2} \right)&=\pm ... the equation ${{z}^{3}}=-8$\\ ====Solution==== \begin{align}{{z}^{3}}&=-8\\ {{z}^{3}}+{{2}^{3}}&=0\\ \l... ,\,\,\,b=-2$ and $c=4$ \\ Quadratic formula is\\ \begin{align}z&=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}\\
- Question, Exercise 10.1 @fsc-part1-kpk:sol:unit10
- 3rd quadrant and $\cos$ is -ive in 3rd quadrant, \begin{align}\cos\alpha &=-\sqrt{1-\sin^2\alpha}\\ &=-\... 2nd quadrant and $\sin$ is +ive in 2nd quadrant, \begin{align}&=\sqrt{1-{{\left(-\dfrac{12}{13} \right)}^... \quad \sin \beta &=\frac{5}{13}.\end{align} Now \begin{align}\sin (\alpha-\beta )&=\sin \alpha\cos \beta... 3rd quadrant and $\cos$ is -ive in 3rd quadrant, \begin{align}\cos\alpha &=-\sqrt{1-\sin^2\alpha}\\ &=-\
- Question 2, Exercise 1.2 @fsc-part1-kpk:sol:unit01
- , that is, $$(z_1+z_2)+z_3=z_1+(z_2+z_3).$$ Take \begin{align} {{z}_{1}}+{{z}_{2}}&=\left( -1+i \right)+\left( 3-2i \right)\\ &=2-i\end{align} So \begin{align} \left( {{z}_{1}}+{{z}_{2}} \right)+{{z}_{3... 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}}+{{z}_{3}} \right
- Question 2, Exercise 1.3 @fsc-part1-kpk:sol:unit01
- is a factor of $P(z)$ iff $P(a)=0$. Put $z=-2$ \begin{align} P(-2)&=(-2)^3+6(-2)+20\\ &=-8-12+20\\ &=0\... 6z+20$.\\ By using synthetic division, we have $$\begin{array}{c|cccc} -2 & 1 & 0 & 6 & 20 \\ & \downa... & 1 & -2 & 10 & 0 \\ \end{array}$$ This gives \begin{align} P(z)&=(z+2)(z^2-2z+10)\\ &=(z+2)\left(z^2... near factors. $$P(z)=3z^2+7.$$ ====Solution==== \begin{align} P(z)&=3z^2+7\\ &=\left(\sqrt{3}z\right)^2-
- Question 2, Exercise 10.2 @fsc-part1-kpk:sol:unit10
- the second quadrant, value of $\cos$ is negative \begin{align}\cos\theta &=-\sqrt{1-{{\sin }^{2}}\theta }... the following by using double angle identities: \begin{align}\sin 2\theta &=2\sin \theta \cos \theta \\ ... the second quadrant, value of $\cos$ is negative \begin{align}\cos\theta &=-\sqrt{1-{{\sin }^{2}}\theta }... the following by using double angle identities: \begin{align}\cos 2\theta &={{\cos }^{2}}\theta -{{\sin
- Question 8, Exercise 1.2 @fsc-part1-kpk:sol:unit01
- ==== Assume $z=a+ib$, then $\overline{z}=a-ib$. \begin{align}z+\overline{z}&=\left( a+ib \right)+\left( ... Assume that $z=a+ib$, then $\overline{z}=a-ib$. \begin{align}z-\overline{z}&=\left( a+ib \right)-\left( ... Suppose $z=a+ib$, then $\overline{z}=a-ib$. Then \begin{align}z\overline{z}&=\left( a+ib \right)\cdot \le... e{z}&={{a}^{2}}+b^2. \ldots (1) \end{align} Now \begin{align} {{\left[\operatorname{Re}\left( z \right)
- Question 2, Exercise 10.1 @fsc-part1-kpk:sol:unit10
- pi }{3}-\dfrac{\pi }{4}$ and using the identity: \begin{align}\sin (\alpha -\beta )=\sin \alpha \cos \beta -\cos \alpha \sin.\end{align} \begin{align} \Rightarrow \quad \sin \left( \frac{\pi }{... irc }}+{{30}^{\circ }}$ and using the identity: \begin{align}\tan (\alpha +\beta )&=\dfrac{\tan \alpha +... c }}+{{45}^{{}^\circ }}$ and using the identity: \begin{align}\tan ({{60}^{\circ }}+{{45}^{\circ }})&=\df
- Question11 and 12, Exercise 10.1 @fsc-part1-kpk:sol:unit10
- $ and $\gamma$ are angles of triangle, therefore \begin{align}&\alpha+\beta +\gamma =180^\circ\\ \implie... ta}{2}=90^\circ-\dfrac{\gamma}{2}\end{align} Now \begin{align}&\tan\left( \dfrac{\alpha }{2}+\dfrac{\beta... amma}{2}\right)=\cot\tfrac{\gamma}{2}\end{align} \begin{align} \implies & \frac{\tan\dfrac{\alpha}{2}\tan... and $\gamma $ are angles of triangle, therefore \begin{align}&\alpha +\beta +\gamma =180^\circ\\ \impli