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- Question 6, Exercise 10.2 @fsc-part1-kpk:sol:unit10
- Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB)... {\circ }}}{2}$, we can find $\cos {{15}^{\circ }}$by using half angle identity as, \begin{align}\cos {... c }}}{2}$, so we can find $\tan {{67.5}^{\circ }}$by using half angle identity as, \begin{align}\tan {... rc }}}{2}$, so we can find $sin{{112.5}^{\circ }}$by using half angle identity as, \begin{align} sin{
- Question 2, Exercise 1.3 @fsc-part1-kpk:sol:unit01
- Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB)... == Given: $$p\left( z \right)={{z}^{3}}+6z+20$$ By factor theorem, $(z-a)$ is a factor of $P(z)$ iff... } Thus $z+2$ is a factor of ${{z}^{3}}+6z+20$.\\ By using synthetic division, we have $$\begin{array}... $$P\left( z \right)={{z}^{3}}-2{{z}^{2}}+z-2$$ \\ By factor theorem $(z-a)$ is a factor of $()$ iff $P
- Question 2, Exercise 10.2 @fsc-part1-kpk:sol:unit10
- Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB)... a = -\dfrac{12}{13}$$ Thus, we have the following by using double angle identities: \begin{align}\sin ... a = -\dfrac{12}{13}$$ Thus, we have the following by using double angle identities: \begin{align}\cos ... a = -\dfrac{12}{13}$$ Thus, we have the following by using double angle identities: Thus, we have the
- Question 3, Exercise 10.2 @fsc-part1-kpk:sol:unit10
- Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB)... al ray of $\theta$ is in the second quadrant and by drawing the reference triangle as shown: {{ :fsc... eta =-\dfrac{3}{5}$. Thus, we have the following by using double angle identity: \begin{align}\sin 2\... al ray of $\theta$ is in the second quadrant and by drawing the reference triangle as shown: {{ :fsc
- Question 5, Exercise 10.3 @fsc-part1-kpk:sol:unit10
- Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB)... 40^\circ) \right)\sin {{20}^{\circ }} \quad \text{by using (1)}\\ &=-\dfrac{\sqrt{3}}{4}\left( \cos {{... circ)-\sin(40^\circ-20^\circ) \right) \quad \text{by using (2)}\\ &=\dfrac{\sqrt{3}}{8}\sin {{20}^{\ci... rc }}-{{10}^{\circ }} \right) \right) \quad \text{by using (1)}\\ &=-\dfrac{1}{4}\sin {{50}^{\circ }}\
- Question 1, Exercise 1.3 @fsc-part1-kpk:sol:unit01
- Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB)... …(i)\\ 2z+3w&=11-5i …(ii)\end{align} Multiply $2$ by (i), we get\\ \begin{align}2z-8w&=6i …(iii)\end{... i …(i)\\ 2z+3w&=2 …(ii)\end{align} Multiply $2$ by (i), we get\\ \begin{align}2z+2w&=6i …(iii)\end{... i …(ii)\end{align} Multiply $\left( 2-i \right)$ by (ii), we get,\\ \begin{align}\left( 2-i \right)\l
- Question 4 and 5, Exercise 10.2 @fsc-part1-kpk:sol:unit10
- Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB)... s gives $\frac{\theta}{2}$ lies in 2nd quadrant. By using the half angle identity: $$\sin\dfrac{\thet... ===Solution==== Given: $\sin \dfrac{2\pi }{3}$.\\ By using double angle identities, we have \begin{ali... ===Solution==== Given: $\cos \dfrac{2\pi }{3}$ \\ By using double angle identities, we have \begin{ali
- Question 7, Exercise 10.2 @fsc-part1-kpk:sol:unit10
- Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB)... t)\left( 1 \right)\\ &=\cos 2\theta \quad \text{(By using double angle identity)}\\ &=\dfrac{1}{\sec ... heta }{2}}\\ &=\dfrac{2}{2\sin \theta }\quad (By \,using\, double\, angle\, identity)\end{align} ... lpha }{2}}{2{{\cos }^{2}}\dfrac{\alpha }{2}}\quad(By\, using\, half\, angle\, identity\, and\, double\
- Unit 1: Complex Numbers (Solutions)
- a first unit of the book Mathematics 11 published by Khyber Pakhtunkhwa Textbook Board, Peshawar, Paki... to * Recall complex number $z$ represented by an expression of the form $z=a+ib$ or of the form... quadratic equation of the form $pz^2+ qz+ r = 0$ by completing squares, where $p,q,r$ are real number
- Question 7, Exercise 1.2 @fsc-part1-kpk:sol:unit01
- Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB)... {2+3i}{5-2i}\times \dfrac{5+2i}{5+2i} \quad \text{by rationalizing} \\ =&\dfrac{10-6+15i+4i}{25+4}\\ =... -3+4i}{1-3i}\times \dfrac{1+3i}{1+3i} \quad \text{by rationalizing}\\ =&\dfrac{-3-12+4i-9i}{1+9}\\ =&
- Question 1, Exercise 10.2 @fsc-part1-kpk:sol:unit10
- Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB)... 1}{5}$, $\theta$ in quadrant II. ====Solution==== By drawing the reference triangle as shown: {{ :fsc... frac{-5}{\sqrt{26}}$ Thus, we have the following by using double angle identities. \begin{align}\sin
- Question 5, Exercise 1.2 @fsc-part1-kpk:sol:unit01
- Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB)... 2a-3bi}\times \dfrac{2a+3bi}{2a+3bi} \quad (\text{by rationalizing})\\ &=\dfrac{-2{{a}^{2}}+9{{b}^{2}}
- Question 6, Exercise 1.2 @fsc-part1-kpk:sol:unit01
- Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB)... &=|{{z}_{1}}|\cdot \dfrac{1}{|{{z}_{2}}|}, \text{ by using (1)}\\ &=\dfrac{|{{z}_{2}}|}{|{{z}_{2}}|}=R
- Question 8, Exercise 1.2 @fsc-part1-kpk:sol:unit01
- Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB)... e{z}=-bi \\ \implies &\overline{z}=-z \quad \text{by using (2).} \end{align} This was required.
- Question11 and 12, Exercise 10.1 @fsc-part1-kpk:sol:unit10
- Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB)... ta \tan \gamma. \end{align} Dividing through out by $\tan \alpha \tan \beta \tan \gamma $ to get \beg