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- Question 7, Exercise 10.2 @math-11-kpk:sol:unit10
- ntity ${{\cos }^{4}}\theta -{{\sin }^{4}}\theta =\dfrac{1}{\sec 2\theta }$. ====Solution==== \begin{align... quad \text{(By using double angle identity)}\\ &=\dfrac{1}{\sec 2\theta }=R.H.S.\end{align} =====Question 7(ii)===== Prove the identity $\tan \dfrac{\theta }{2}+co\operatorname{t}\dfrac{\theta }{2}=\dfrac{2}{\sin \theta }$. ====Solution==== \begin{align
- Question 1 Exercise 5.3 @math-11-kpk:sol:unit05
- ===Question 1(i)==== Find the sum of the series $\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+\ldots$ to $n$ terms. ====Solution==== The general term of the series is: $$T_n=\dfrac{1}{n(n+1)}$$ Resolving $T_n$ into partial fractio
- Question 2, Exercise 2.3 @math-11-kpk:sol:unit02
- eft| \begin{matrix} -3 & 4 & -5 \\ 1 & -1 & 2 \\ \dfrac{5}{49} & -\dfrac{6}{49} & \dfrac{8}{49}\end{matrix} \right. \right]\text{by}-\dfrac{1}{49}R_3\\ \underset{\sim}{R}&\left[\begin{matrix} 1
- Question 4 Review Exercise @math-11-kpk:sol:unit05
- Pakistan. =====Question 4===== Sum the series: $\dfrac{1}{1.4 .7}+\dfrac{1}{4.7 .10}+\dfrac{1}{7.10 .13}+\ldots$ ====Solution==== In the denominator Each term is the product of ... $ Thus the general term of the series is: $$a_n=\dfrac{1}{(3 n-2)(3 n+1)(3 n+4)}$$ Resolving into partia
- Question 2, Exercise 10.1 @math-11-kpk:sol:unit10
- ====Question 2(i)==== Evaluate exactly: $\sin \dfrac{\pi }{12}$ ===Solution=== We rewrite $\dfrac{\pi }{12}$ as $\dfrac{\pi }{3}-\dfrac{\pi }{4}$ and using the identity: \begin{align}\sin (\alpha -\beta )=\sin \alpha \cos \b
- Question 5, Exercise 10.3 @math-11-kpk:sol:unit10
- \circ }}\cos {{60}^{\circ }}\cos {{80}^{\circ }}=\dfrac{1}{16}.$$ ====Solution==== We know that\\ $2\cos ... &=\cos {{20}^{\circ }}\cos {{40}^{\circ }}\left( \dfrac{1}{2} \right)\cos {{80}^{\circ }}\\ &=\dfrac{1}{2}\cos {{80}^{\circ }}\cos {{40}^{\circ }}\cos {{20}^{\circ }}\\ &=\dfrac{1}{4}\left( 2\,\cos {{80}^{\circ }}\cos {{40}^{\c
- Question 5, Exercise 10.3 @math-11-kpk:sol:unit10
- \circ }}\cos {{60}^{\circ }}\cos {{80}^{\circ }}=\dfrac{1}{16}$. ====Solution==== We know that\\ $2\cos \... &=\cos {{20}^{\circ }}\cos {{40}^{\circ }}\left( \dfrac{1}{2} \right)\cos {{80}^{\circ }}\\ &=\dfrac{1}{2}\cos {{80}^{\circ }}\cos {{40}^{\circ }}\cos {{20}^{\circ }}\\ &=\dfrac{1}{4}\left( 2\,\cos {{80}^{\circ }}\cos {{40}^{\c
- Question 6, Exercise 10.2 @math-11-kpk:sol:unit10
- c }}$. ====Solution==== Because ${{15}^{\circ }}=\dfrac{{{30}^{\circ }}}{2}$, and $\dfrac{\theta }{2}=\dfrac{{{30}^{\circ }}}{2}$, we can find $\cos {{15}^{\circ }}$by using half angle identity as, \begin{align}\cos {{15}^{\circ }}&=\cos \dfrac{{{30}^{\circ }}}{2}=\sqrt{\dfrac{1+\cos {{30}^{\c
- Question 3, Exercise 10.1 @math-11-kpk:sol:unit10
- ar, Pakistan. =====Question 3(i)===== If $\sin u=\dfrac{3}{5}$ and $\sin v=\dfrac{4}{5}$ where$u$ and $v$ are between $0$ and $\dfrac{\pi }{2}$, evaluate each of the following exactly ... ft( u+v \right)$ ====Solution==== Given $\sin u=\dfrac{3}{5},$ $0\le u\le \dfrac{\pi }{2}.$ $\sin v
- Question 7, Exercise 1.2 @math-11-kpk:sol:unit01
- (i)===== Separate into real and imaginary parts $\dfrac{2+3i}{5-2i}$. ====Solution==== \begin{align}&\dfrac{2+3i}{5-2i} \\ =&\dfrac{2+3i}{5-2i}\times \dfrac{5+2i}{5+2i} \quad \text{by rationalizing} \\ =&\dfrac{10-6+15i+4i}{25+4}\\ =&\d
- Question 8, Exercise 10.1 @math-11-kpk:sol:unit10
- ===Question 8(i)===== Prove that: $\tan \left( \dfrac{\pi }{4}+\theta \right)=\dfrac{\cos \theta +\sin \theta }{\cos \theta -\sin \theta }$ ====Solution==== \begin{align}L.H.S.&=\tan \left( \dfrac{\pi }{4}+\theta \right)\\ &=\dfrac{\sin \left( \dfrac{\pi }{4}+\theta \right)}{\cos \left( \dfrac{\pi
- Question 1 Exercise 4.5 @math-11-kpk:sol:unit04
- = In the given geometric series: $a_1=3, \quad r=\dfrac{6}{3}=2$ and $a_n=3.2^9$.\\ We first find $n$ and... in{align}3.2^9&=3(2)^{n-1} \text { or }(2)^{n-1}=\dfrac{3.2^9}{3} \\ \Rightarrow(2)^{n-1}&=2^9 \\ \Righta... 9 \text { or } n=10 \\ \text {. Now }\quad S_n&=\dfrac{a_1(r^n-1)}{r-1},\end{align} becomes in the given case\\ \begin{align}S_{10}&=\dfrac{3[2^{10}-1]}{2-1} \\ \Rightarrow \quad S_{10}&=3(
- Question 9 Exercise 4.4 @math-11-kpk:sol:unit04
- 9(i)===== Insert five geometric means between $3 \dfrac{5}{9}=\dfrac{32}{9}\quad$ and $\quad40 \dfrac{1}{2}=\dfrac{81}{2}$. ====Solution==== Let $G_1, G_2, G_3, G_4$ and $G_5$ be the five geometric means be
- Question, Exercise 10.1 @math-11-kpk:sol:unit10
- stan. =====Question 4(i)===== If $\sin \alpha =-\dfrac{4}{5}$ and $\cos \beta =-\dfrac{12}{13}$, $\alpha $in Quadrant III and $\beta $in Quadrant II, find th... \right)$. ====Solution==== Given: $\sin \alpha=-\dfrac{4}{5}$, $\alpha$ is in 3rd quadrant, \\ $\sin \beta=-\dfrac{12}{13}$, $\beta$ is in 2nd quadrant. We have an
- Question11 and 12, Exercise 10.1 @math-11-kpk:sol:unit10
- the angles of a triangle $ABC$, show that $\cot \dfrac{\alpha }{2}+\cot \dfrac{\beta }{2}+\cot \dfrac{\gamma }{2}=\cot \dfrac{\alpha }{2}\cot \dfrac{\beta }{2}\cot \dfrac{\gamma }{2}$ ====Solution==== Sinc