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- Question 5, Exercise 10.1 @fsc-part1-kpk:sol:unit10
- nt, therefor it lies in 3rd quadrant. Now \begin{align}{{\sec}^{2}}\alpha &=1+{{\tan}^{2}}\alpha\\ \Righ... sec \alpha &=\pm \sqrt{1+{{\tan}^{2}}\alpha}.\end{align} Since terminal arm of $\alpha$ is in the 3rd quadrant, value of $\sec$ is –ive \begin{align}\sec \alpha &=-\sqrt{1+{{\tan}^{2}}\alpha }\\ \Ri... \Rightarrow \quad \sec \alpha&=-\frac{5}{4} \end{align} Now $$\cos\alpha =\frac{1}{\sec \alpha }=\frac{
- Question 13, Exercise 10.1 @fsc-part1-kpk:sol:unit10
- nd $3=r\sin\varphi$. Squaring and adding \begin{align} &r^2\cos^2 \varphi+r^2\sin^2\varphi = 4^2+3^2\\ ... = 16+9\\ \implies &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
- Question 7, Exercise 10.2 @fsc-part1-kpk:sol:unit10
- dfrac{1}{\sec 2\theta }$. ====Solution==== \begin{align}L.H.S&={{\cos }^{4}}\theta -{{\sin }^{4}}\theta \... dentity)}\\ &=\dfrac{1}{\sec 2\theta }=R.H.S.\end{align} =====Question 7(ii)===== Prove the identity $\t... \dfrac{2}{\sin \theta }$. ====Solution==== \begin{align}L.H.S&=\tan \dfrac{\theta }{2}+co\operatorname{t}... (By \,using\, double\, angle\, identity)\end{align} =====Question 7(iii)===== Prove the identity $\
- Question 3, Exercise 10.1 @fsc-part1-kpk:sol:unit10
- t and $\cos $ is $+ve$ in first quadrant . \begin{align}\Rightarrow \,\,\,\,\,\cos u&=\sqrt{1-{{\sin }^{2... \\ \Rightarrow \,\,\,\,\,\cos u&=\dfrac{4}{5}\end{align} Also $\cos v=\sqrt{1-{{\sin }^{2}}v}$ As $v$ lie... nt and $\cos $ is $+ve$ in first quadrant. \begin{align}\Rightarrow \,\,\,\,\,\cos v&=\sqrt{1-{{\sin }^{2... 25}}\\ \Rightarrow \,\,\,\cos v &=\frac{3}{5}\end{align} Now \begin{align}\cos \left( u+v \right)&=\cos
- Question 1, Exercise 1.3 @fsc-part1-kpk:sol:unit01
- 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 …(i
- Question 1, Exercise 10.1 @fsc-part1-kpk:sol:unit10
- n {{22}^{\circ }}$ ==== Solution ==== As \begin{align} \sin (\alpha +\beta )=\sin \alpha \cos \beta +\cos \alpha \sin \beta, \end{align} Therefore \begin{align} \sin {{37}^{\circ }}\cos {{22}^{\circ }}+\cos {{37}^{\circ }}\sin {{22}^{\cir... t( 37+22 \right) \\ & =\sin {{59}^{\circ }}. \end{align} ===== Question 1(ii)===== Write as a trigonom
- Question 5, Exercise 1.2 @fsc-part1-kpk:sol:unit01
- -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\
- Question, Exercise 10.1 @fsc-part1-kpk:sol:unit10
- adrant and $\cos$ is -ive in 3rd quadrant, \begin{align}\cos\alpha &=-\sqrt{1-\sin^2\alpha}\\ &=-\sqrt{1... \Rightarrow \quad \cos \alpha&=-\dfrac{3}{5}.\end{align} Also $$\sin \beta=\pm\sqrt{1-\cos^2\beta}.$$ As ... adrant and $\sin$ is +ive in 2nd quadrant, \begin{align}&=\sqrt{1-{{\left(-\dfrac{12}{13} \right)}^{2}}}\... \Rightarrow \quad \sin \beta &=\frac{5}{13}.\end{align} Now \begin{align}\sin (\alpha-\beta )&=\sin \al
- Question 6, Exercise 1.3 @fsc-part1-kpk:sol:unit01
- z}^{4}}+{{z}^{2}}+1=0$\\ ====Solution==== \begin{align}{{z}^{4}}+{{z}^{2}}+1&=0\\ {{z}^{4}}+2\left( \dfr... 2}}+\dfrac{1}{2} \right)}^{2}}&=-\dfrac{3}{4}\end{align}\\ Take square root on both sides.\\ \begin{align}\left( {{z}^{2}}+\dfrac{1}{2} \right)&=\pm \sqrt{-\dfr... \dfrac{\sqrt{3}}{2}i \right)}^{\dfrac{1}{2}}}\end{align} =====Question 6(ii)===== Find the solutions of
- Question 2, Exercise 1.2 @fsc-part1-kpk:sol:unit01
- is, $$(z_1+z_2)+z_3=z_1+(z_2+z_3).$$ Take \begin{align} {{z}_{1}}+{{z}_{2}}&=\left( -1+i \right)+\left( 3-2i \right)\\ &=2-i\end{align} So \begin{align} \left( {{z}_{1}}+{{z}_{2}} \right)+{{z}_{3}}&=\left( 2-i \right)+\left( 2-2i \right)\\ &=4-3i \ldots (1)\end{align} Now \begin{align} {{z}_{2}}+{{z}_{3}}&=\left( 3-
- Question 2, Exercise 10.2 @fsc-part1-kpk:sol:unit10
- cond quadrant, value of $\cos$ is negative \begin{align}\cos\theta &=-\sqrt{1-{{\sin }^{2}}\theta }\\ &=-... ac{5}{13}\right)}\\ &=-\sqrt{\frac{144}{169}}\end{align} $$\implies \cos\theta = -\dfrac{12}{13}$$ Thus, ... ollowing by using double angle identities: \begin{align}\sin 2\theta &=2\sin \theta \cos \theta \\ &=2\le... c{5}{13} \right)\left( \dfrac{12}{13} \right)\end{align} $$\implies \bbox[4px,border:2px solid black]{\si
- 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( a-ib \... erline{z}&=2\operatorname{Re}\left( z \right)\end{align} =====Question 8(ii)===== Show that $z-\overlin... e that $z=a+ib$, then $\overline{z}=a-ib$. \begin{align}z-\overline{z}&=\left( a+ib \right)-\left( a-ib \... =2bi\\ z-\overline{z}&=2i\operatorname{Im}(z)\end{align} =====Question 8(iii)===== Show that $z\overlin
- Question 2, Exercise 10.1 @fsc-part1-kpk:sol:unit10
- }-\dfrac{\pi }{4}$ 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 }{3}-\frac{\pi }{4} \right) & =\sin \frac{\pi }{... qrt{2}}{4} \\ &=\frac{\sqrt{6}-\sqrt{2}}{4}. \end{align} ===Question 2(ii)=== Evaluate exactly:$\tan {
- Question11 and 12, Exercise 10.1 @fsc-part1-kpk:sol:unit10
- $\gamma$ are angles of triangle, therefore \begin{align}&\alpha+\beta +\gamma =180^\circ\\ \implies &\al... }+\dfrac{\beta}{2}=90^\circ-\dfrac{\gamma}{2}\end{align} Now \begin{align}&\tan\left( \dfrac{\alpha }{2}+\dfrac{\beta }{2}\right)=\tan \left( 90-\dfrac{\gamma }... tfrac{\gamma}{2}\right)=\cot\tfrac{\gamma}{2}\end{align} \begin{align} \implies & \frac{\tan\dfrac{\alpha
- Question 7, Exercise 1.1 @fsc-part1-kpk:sol:unit01
- know that $z_1=1+2i$ and $z_2=2+3i$, then \begin{align} {{z}_{1}}+{{z}_{2}}&=1+2i+2+3i\\ &=1+2+2i+3i\\ &=3+5i \end{align} Now \begin{align} |z_1+z_2|&=\sqrt{3^2+5^2}\\ &=\sqrt{9+25}\\ &=\sqrt{34}\end{align} =====Question 7(ii)===== If ${{z}_{1}}=1+2i$