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- Question 5, Exercise 10.1 @math-11-kpk:sol:unit10
- 5(i)===== If $\tan \alpha =\dfrac{3}{4}$, $\sec \beta =\dfrac{13}{5}$ and neither the terminal side of the angle of measure $\alpha$ nor $\beta$ in the first Quadrant, then find: $\sin \left( \alpha +\beta \right)$. ====Solution==== Given: $\tan\alpha... quadrant, therefor it lies in 3rd quadrant. Now \begin{align}{{\sec}^{2}}\alpha &=1+{{\tan}^{2}}\alph
- Question 2, Exercise 2.3 @math-11-kpk:sol:unit02
- the matrix by using elementary row operation. $$\begin{bmatrix}4 & -2 & 5 \\ 2 & 1 & 0 \\ -1 & 2 & 3 \end{bmatrix}$$ ====Solution==== Let $$A=\begin{bmatrix} 4 & -2 & 5 \\ 2 & 1 & 0 \\ -1 & 2 & 3 \end{bmatrix}.$$ Then \begin{align}|A|&=\begin{vmatrix}4 & -2 & 5 \\ 2 & 1 & 0 \\ -1 & 2 & 3 \end{vmatrix}\\ &=4(3)+2(6)+5(4+
- Question 3, Exercise 2.1 @math-11-kpk:sol:unit02
- shawar, Pakistan. =====Question 3(i)===== If $A=\begin{bmatrix}x & y & z\end{bmatrix}$, $B=\begin{bmatrix}a & h & g\\h & b & f\\g & f & c\end{bmatrix}$ and $C=\begin{bmatrix}x\\y\\z\end{bmatrix}$. Verify that $\l... =A\left( BC \right)$. ====Solution==== Given: $A=\begin{bmatrix}x & y & z\end{bmatrix}$, $B=\begin{bma
- Question11 and 12, Exercise 10.1 @math-11-kpk:sol:unit10
- r, Pakistan. =====Question 11===== If $\alpha$, $\beta$, $\gamma$ are 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==== Since $\alpha$, $\beta$ and $\gamma$ are angles of triangle, therefore
- Question, Exercise 10.1 @math-11-kpk:sol:unit10
- ===== If $\sin \alpha =-\dfrac{4}{5}$ and $\cos \beta =-\dfrac{12}{13}$, $\alpha $in Quadrant III and $\beta $in Quadrant II, find the exact value of $\sin \left( \alpha -\beta \right)$. ====Solution==== Given: $\sin \alpha... c{4}{5}$, $\alpha$ is in 3rd quadrant, \\ $\sin \beta=-\dfrac{12}{13}$, $\beta$ is in 2nd quadrant. W
- Question 1, Exercise 2.1 @math-11-kpk:sol:unit02
- n 1(i)===== Express as a single matrix $$\left[ \begin{matrix} 1 & 2 & 4 \\ \end{matrix} \right] \left[ \begin{matrix} 1 & 0 & 2 \\ 2 & 0 & 1 \\ 0 & 1 & 2 \\ \end{matrix} \right] \left[ \begin{matrix} 2 \\ 4 \\ 6 \\ \end{matrix} \right]$$ ====Solution==== \begin{align}&\left[ \begin{matrix} 1 & 2 & 4 \\
- Question 13 Exercise 6.2 @math-11-kpk:sol:unit06
- word "Excellence." How many of these permutations begin with $\mathrm{E}$ ? ====Solution==== The total... m_2=2$ are $L$ and $m_3=2$ are $C$. Therefore, \begin{align}\text{total number of permutations are} &=\left(\begin{array}{c} n \\ m_1, m_2, m_3 \end{array}\right)\\&=\left(\begin{array}{c} 10 \\ 4,2,2 \end{array}\right) \\ &
- Question 7, Exercise 10.2 @math-11-kpk:sol:unit10
- eta =\dfrac{1}{\sec 2\theta }$. ====Solution==== \begin{align}L.H.S&={{\cos }^{4}}\theta -{{\sin }^{4}... }{2}=\dfrac{2}{\sin \theta }$. ====Solution==== \begin{align}L.H.S&=\tan \dfrac{\theta }{2}+co\operat... \theta }{2}\cos \dfrac{\theta }{2}} \quad \because \left( {{\sin }^{2}}\theta +{{\cos }^{2}}\th... \theta }={{\cot }^{2}}\theta $. ====Solution==== \begin{align}L.H.S&=\dfrac{1+\cos 2\theta }{1+\cos 2\
- Question 10 Exercise 7.2 @math-11-kpk:sol:unit07
- $. Solution: We know that $$ \left.(1+x)^n=\left(\begin{array}{l} n \\ \vdots \end{array}\right)+\left(\begin{array}{l} m \\ 1 \end{array}\right) x+\left(\begin{array}{l} n \\ 2 \end{array}\right) x^2-\ldots+... the above equation, we have $(1 \div 1)^n=\left(\begin{array}{l}n \\ 0\end{array}\right)+\left(\begin
- Question 7, Exercise 10.1 @math-11-kpk:sol:unit10
- tion 7(i)===== Show that: $\cot \left( \alpha +\beta \right)=\dfrac{\cot \alpha \cot \beta -1}{\cot \alpha +\cot \beta }$ ====Solution==== \begin{align}L.H.S.&=\cot (\alpha +\beta )\\ &=\dfrac{1}{\tan (\alpha +\beta )
- Question 1, Exercise 2.3 @math-11-kpk:sol:unit02
- )===== Reduce the matrices to the echelon form: $\begin{bmatrix}1 & 3 & -1 \\2 & 1 & 4 \\3 & 4 & -5\end{bmatrix}$. ====Solution==== \begin{align}&\begin{bmatrix} 1 & 3 & -1 \\ 2 & 1 & 4 \\ 3 & 4 & -5 \end{bmatrix}\\ \underset{\sim}{R}&\begin{bmatrix} 1 & 3 & -1 \\ 0 & -5 & 6 \\ 0 & -5
- Question 10 Exercise 7.1 @math-11-kpk:sol:unit07
- an. =====Question 10===== Establish the formulas below by mathematical induction, $\left(\begin{array}{1}5 \\5 \end{array}\right)+\left(\begin{array}{l}6 \\ 5\end{array}\right)+\left(\begin{array}{l}7 \\ 5\end{array}\right)+\ldots+\left(\beg
- Question 11 Exercise 7.1 @math-11-kpk:sol:unit07
- PTBB) Peshawar, Pakistan. =====Question 11===== \begin{align} & \left(\begin{array}{l} 2 \\ 2 \end{array}\right)+\left(\begin{array}{l} 3 \\ 2 \end{array}\right)+\left(\begin{array}{l} 4 \\ 2 \end{array}\right)+\ldots+\left(
- Question 9, Exercise 2.1 @math-11-kpk:sol:unit02
- shawar, Pakistan. =====Question 9(i)===== If $A=\begin{bmatrix}2 & -1 & 3 \\1 & \quad 0 & 1 \end{bmatrix},$ $B=\begin{bmatrix}1 & 2 \\2 & 2 \\ 3 & 0 \end{bmatrix}... $( AB )^t=B^tA^t$. ====Solution==== $$A=\left[ \begin{matrix} 2 & -1 & 3 \\ 1 & \quad 0 & 1 \\ \end{matrix} \right],$$ $$B=\left[ \begin{matrix} 1 & 2 \\ 2 & 2 \\ 3 & 0 \\
- Question 9 Exercise 6.3 @math-11-kpk:sol:unit06
- Question 9(i)===== An $8$-persons committee is to be formed from a group of $6$ women and $7$ men. In how many ways can the committee be chosen if (i) committee must contain four men and... Total number of different ways that four men to be selected are: ${ }^7 C_4$. Total number of different ways that four women to be selected are ${ }^6 C_4$. By fundamental princip