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- Question, Exercise 10.1 @fsc-part1-kpk:sol:unit10
- n \alpha =-\dfrac{4}{5}$ and $\cos \beta =-\dfrac{12}{13}$, $\alpha $in Quadrant III and $\beta $in ... lpha$ is in 3rd quadrant, \\ $\sin \beta=-\dfrac{12}{13}$, $\beta$ is in 2nd quadrant. We have an ide... quadrant, \begin{align}&=\sqrt{1-{{\left(-\dfrac{12}{13} \right)}^{2}}}\\ &=\sqrt{1-\dfrac{144}{169}}... \beta \\ &=\left(-\frac{4}{5}\right)\left(-\frac{12}{13}\right)+\left(-\frac{3}{5}\right)\left(\frac{
- Question 2, Exercise 10.1 @fsc-part1-kpk:sol:unit10
- on 2(i)==== Evaluate exactly: $\sin \dfrac{\pi }{12}$ ===Solution=== We rewrite $\dfrac{\pi }{12}$ as $\dfrac{\pi }{3}-\dfrac{\pi }{4}$ and using the ide... ion 2(iv)=== Evaluate exactly:$\tan \dfrac{5\pi }{12}$ ==Solution== We rewrite $\dfrac{5\pi }{12}$ as $\dfrac{\left( 2+3 \right)\pi }{12}$ or $\dfrac{\pi
- Question 3, Exercise 10.1 @fsc-part1-kpk:sol:unit10
- 5}.\frac{3}{5}-\frac{3}{5}.\frac{4}{5}\\ &=\dfrac{12}{25}-\frac{12}{25}\\ \cos \left( u+v \right)&=0\end{align} =====Question 3(ii)===== If $\sin u=\df... dfrac{3}{4}\dfrac{4}{3}}\\ &=\dfrac{\dfrac{9-16}{12}}{1+\dfrac{12}{12}}\\ &=\dfrac{\dfrac{-7}{12}}{2}\\ &=\dfrac{-7}{24}\end{align} =====Question 3(iii
- Question 5, Exercise 10.1 @fsc-part1-kpk:sol:unit10
- 4}{169}} \\ \Rightarrow \quad \sin\beta&=-\dfrac{12}{13} \end{align} Now \begin{align} \sin(\alph... {13}\right)+\left(-\frac{4}{5}\right)\left(-\frac{12}{13}\right)\\ &=-\frac{3}{13}+\frac{48}{65}\end{a... 4}{169}} \\ \Rightarrow \quad \sin\beta&=-\dfrac{12}{13} \end{align} Now \begin{align} \cos(\alph... }{13}\right)-\left(\frac{3}{5}\right)\left(-\frac{12}{13}\right)\\ &=-\frac{20}{65}+\frac{36}{65}\end{
- Question 2, Exercise 10.2 @fsc-part1-kpk:sol:unit10
- {169}}\end{align} $$\implies \cos\theta = -\dfrac{12}{13}$$ Thus, we have the following by using doubl... \\ &=2\left( -\dfrac{5}{13} \right)\left( \dfrac{12}{13} \right)\end{align} $$\implies \bbox[4px,border:2px solid black]{\sin 2\theta=-\dfrac{120}{169}.}$$ =====Question 2(ii)===== If $\sin \t... {169}}\end{align} $$\implies \cos\theta = -\dfrac{12}{13}$$ Thus, we have the following by using doubl
- Question 9 & 10, Exercise 1.1 @fsc-part1-kpk:sol:unit01
- ight)}\\ &=\dfrac{6+6+9i-4i}{2+2+4i-i}\\ &=\dfrac{12+5i}{4+3i}\end{align} Now \begin{align}\bar{z}&=\dfrac{12-5i}{4-3i}\\ &=\dfrac{12-5i}{4-3i}\times \dfrac{4+3i}{4+3i}\\ &=\dfrac{48+15+36i-20i}{16+9+12i-12i}\\ ... t)}^{9}}+\dfrac{1}{i.{{\left( {{i}^{2}} \right)}^{12}}}\\ &={{\left( -1 \right)}^{9}}+\dfrac{1}{i.{{\l
- Question 1, Exercise 10.1 @fsc-part1-kpk:sol:unit10
- single angle. $\dfrac{\tan {{35}^{\circ }}-\tan {{12}^{\circ }}}{1+\tan {{35}^{\circ }}\tan {{12}^{\circ }}}$ ==== Solution ==== As \begin{align}\tan (\a... \begin{align}\dfrac{\tan {{35}^{\circ }}-\tan {{12}^{\circ }}}{1+\tan {{35}^{\circ }}\tan {{12}^{\circ }}}&=\tan \left( 35-12 \right)\\ &=\tan {{23}^{\c
- Question 6, Exercise 10.2 @fsc-part1-kpk:sol:unit10
- half angle identities to evaluate exactly $sin{{112.5}^{\circ }}$. ====Solution==== Because ${{112.5}^{\circ }}=\dfrac{{{225}^{\circ }}}{2}$, the $\dfrac... frac{{{225}^{\circ }}}{2}$, so we can find $sin{{112.5}^{\circ }}$by using half angle identity as, \begin{align} sin{{112.5}^{\circ }}&=\sin \dfrac{{{225}^{\circ }}}{2}=\s
- Question11 and 12, Exercise 10.1 @fsc-part1-kpk:sol:unit10
- ====== Question11 and 12, Exercise 10.1 ====== Solutions of Question 11 and 12 of Exercise 10.1 of Unit 10: Trigonometric Identi... cot\dfrac{\gamma}{2}.\end{align} =====Question 12===== If $\alpha +\beta +\gamma ={{180}^{\circ}}$,
- Question 1, Exercise 10.2 @fsc-part1-kpk:sol:unit10
- ox[4px,border:2px solid black]{\cos2\theta=\dfrac{12}{13}}$$ \begin{align}\tan 2\theta &=\dfrac{\sin 2... heta }{\cos 2\theta }=\dfrac{\frac{-5}{13}}{\frac{12}{13}}\end{align} $$ \implies \bbox[4px,border:2px solid black]{\tan 2\theta=-\dfrac{5}{12}}$$ <text align="right"><btn type="success">[[fs
- Question 2, Exercise 1.3 @fsc-part1-kpk:sol:unit01
- =-2$ \begin{align} P(-2)&=(-2)^3+6(-2)+20\\ &=-8-12+20\\ &=0\end{align} Thus $z+2$ is a factor of ${... =2$\\ \begin{align}P(-2)&=(-2)^3+6(-2)+20\\ &=-8-12+20=0\end{align} Thus $z-2$ is a factor of ${{z}^
- Question 6, Exercise 1.3 @fsc-part1-kpk:sol:unit01
- frac{2\pm \sqrt{4-16}}{2}\\ z&=\dfrac{2\pm \sqrt{-12}}{2}\\ z&=\dfrac{2\pm 2\sqrt{3}i}{2}\\ z&=1\pm \s... ht)}}{2\left( 1 \right)}\\ z&=\dfrac{3\pm \sqrt{6-12}}{2}\\ z&=\dfrac{3\pm \sqrt{-6}}{2}\\ z&=\dfrac{3
- Definitions: FSc Part1 KPK
- Peshawar, Pakistan. The book has total of twelve (12) chapters. Definition of the book provide the qu
- Important Questions
- odha(2016)// * Find the value of $cos\frac{\pi}{12}$ --- // BISE Sargodha(2017)// * Show that $\fr
- Unit 10: Trigonometric Identities of Sum and Difference of Angles (Solutions) @fsc-part1-kpk:sol
- [[fsc-part1-kpk:sol:unit10:ex10-1-p10|Question 11-12]] * [[fsc-part1-kpk:sol:unit10:ex10-1-p11|Quest