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- Question 1, Exercise 8.1 @math-11-nbf:sol:unit08
- on 1(i)===== Find the value of $\cos (\alpha \pm \beta), \sin (\alpha \pm \beta)$ and $\tan (\alpha \pm \beta)$ for the pair of angles. $\alpha=180^{\circ}, \beta=60^{\circ}$ ** Solution. ** Given: $\alpha=18
- Question 11, Exercise 8.1 @math-11-nbf:sol:unit08
- 270^{\circ}-\lambda\right)}=1$ ** Solution. ** \begin{align*} L.H.S & = \dfrac{\sin \left(180^{\circ... (90^{\circ}+\alpha\right)}=-1$ ** Solution. ** \begin{align*} L.H.S & = \frac{\sin \left(90^{\circ}+... estion 11(iii)===== Show that: $\tan \alpha+\tan \beta=\dfrac{\sin (\alpha+\beta)}{\cos \alpha \cos \beta}$ ** Solution. ** \begin{align*} L.H.S & = \ta
- Question 4 Exercise 8.2 @math-11-nbf:sol:unit08
- \theta$ lies in QI, $\sin$ is positive in QI, so \begin{align*} \sin\theta & = \sqrt{1-\left(\frac{3}... = \frac{4}{5} \end{align*} (a) $\sin 2 \theta$ \begin{align*} \sin 2\theta & = 2 \sin\theta \cos\the... \right) \left(\frac{3}{5} \right)\\ \end{align*} \begin{align*} \implies \boxed{\sin 2\theta = \frac{24}{25}}. \end{align*} (b) $\cos 2 \theta$ \begin{align*} \cos 2\theta & = 1-2\sin^2\theta \\ &
- Question 5 and 6, Exercise 8.1 @math-11-nbf:sol:unit08
- stion 5===== For $\sin \alpha=\dfrac{4}{5}, \tan \beta=-\dfrac{5}{12}$ with terminal side of an angles in QII, find $\cos (\alpha+\beta)$ and $\cos (\alpha-\beta)$. ** Solution. ** Given: $\sin \alpha=\dfrac{4}{5}$, $\alpha$ is in QII and $\tan \beta=-\dfrac{5}{12}$, $\beta$ is in QII. We have an
- Question 12, Exercise 8.1 @math-11-nbf:sol:unit08
- akistan. ===== Question 12(i)===== If $\alpha+\beta+\gamma=180^{\circ}$, prove that: $\tan \alpha+\tan \beta+\tan \gamma=\tan \alpha \tan \beta \tan \gamma$. ** Solution. ** Given: $$\alpha+\beta+\gamma=180^{\circ}$$ This gives \begin{align*}
- Question 9, Exercise 8.1 @math-11-nbf:sol:unit08
- . ===== Question 9(i)===== Given $\alpha$ and $\beta$ are obtuse angles with $\sin \alpha=\dfrac{1}{\sqrt{2}}$ and $\cos \beta=-\dfrac{3}{5}$ find: $\sin (\alpha \pm \beta)$ ** Solution. ** Given: $\sin \alpha=\dfrac{1}{\sq... pha$ is obtuse angle, i.e. it is in QII.\\ $\cos \beta=-\dfrac{3}{5}$, $\beta$ is obtuse angle, i.e. i
- Question 1, Exercise 2.5 @math-11-nbf:sol:unit02
- helon form then into reduced echelon form $\left[\begin{array}{ccc}1 & 3 & 5 \\ -6 & 8 & 3 \\ -4 & 6 & 5\end{array}\right]$. ** Solution. ** \begin{align*} & \quad \left[\begin{array}{ccc}1 & 3 & 5 \\ -6 & 8 & 3 \\ -4 & 6 & 5\end{array}\right]\\ \sim & \text{R} \left[\begin{array}{ccc} 1 & 3 & 5 \\ 0 & 26 & 33 \\ 0 & 18
- Question 6, Exercise 2.6 @math-11-nbf:sol:unit02
- lution. ** For this system of equations; we have \begin{align*} A &= \begin{bmatrix} 5 & 3 & 1 \\ 2 & 1 & 3 \\ 1 & 2 & 4 \end{bmatrix}, \quad X = \begin{bmatrix} x \\ y \\ z \end{bmatrix}, \quad B = \begin{bmatrix} 6 \\ 19 \\ 25 \end{bmatrix} \end{alig
- Question 3, Exercise 2.5 @math-11-nbf:sol:unit02
- perations, find the inverse of the matrix $\left[\begin{array}{ccc}0 & -1 & -1 \\ -1 & 3 & 0 \\ 1 & -1... at $A A^{-1}=A^{-1} A=I$.\\ ** Solution. ** Let \begin{align*} A&=\left[ \begin{array}{ccc} 0 & -1 & -1 \\ -1 & 3 & 0 \\ 1 & -1 & 4 \end{array} \rig... end{align*} So $A$ is non singular. Now consider \begin{align*} &\quad\left[ \begin{array}{ccc|ccc} 0
- Question 4, Exercise 2.2 @math-11-nbf:sol:unit02
- Pakistan. =====Question 4(i)===== Find $A$ if \begin{align}\left[\begin{array}{cc} 2 & 1 \\ 3 & 2 \end{array}\right]A\left[\begin{array}{cc} 1 & 3 \\ 2 & 4 \end{array}\right]&=\left[\begin{array}{cc} 1 & 0 \\ 0 & 1 \end{array}\right]\
- Question 5, Exercise 2.3 @math-11-nbf:sol:unit02
- g matrices if it exists by adjoint method $\left[\begin{array}{ccc}1 & -1 & 1 \\ 2 & 1 & -1 \\ 1 & -2 ... -1\end{array}\right]$. ** Solution. ** Given \begin{align*} A &= \left[\begin{array}{ccc} 1 & -1 & 1 \\ 2 & 1 & -1 \\ 1 & -2 & -1 \end{array}\right]\\... gular.\\ Let find the cofactor matrix for $A$.\\ \begin{align*} A_{11} &= (-1)^{1+1} \left|\begin{arra
- Question 5, Exercise 1.4 @math-11-nbf:sol:unit01
- tan. =====Question 5===== If $\cos \alpha+\cos \beta+\cos \gamma=\sin \alpha+\sin \beta+\sin \gamma=0$, show that: (i) $\cos 3 \alpha+\cos 3 \beta+\cos 3 \gamma=3 \cos (\alpha+\beta+\gamma)$. (ii) $\sin 3 \alpha+\sin 3 \beta+\sin 3 \gamma=3 \sin
- Question 7, Exercise 2.3 @math-11-nbf:sol:unit02
- rify that $(A B)^{-1}=B^{-1} A^{-1}$ if $A=\left[\begin{array}{ll}2 & 1 \\ 8 & 6\end{array}\right]$ and $B=\left[\begin{array}{ll}3 & 2 \\ 0 & 2\end{array}\right]$. ** Solution. ** Given: \begin{align*} A &= \left[\begin{array}{ll}2 & 1 \\ 8 & 6\end{array}\right] \\ |A|& = 12 - 8 = 4\\ A^{-
- Question 2, Exercise 2.5 @math-11-nbf:sol:unit02
- ===== Find the rank of each of the matrix $\left[\begin{array}{ccc}5 & 9 & 3 \\ 3 & -5 & 6 \\ 2 & 10 & 6\end{array}\right]$ ** Solution. ** \begin{align*}&\quad\left[ \begin{array}{ccc} 5 & 9 & 3 \\ 3 & -5 & 6 \\ 2 & 10 & 6 \end{array} \right]\\ \sim & \text{R}\left[ \begin{array}{ccc} 1 & \frac{9}{5} & \frac{3}{5} \\
- Question 7, Exercise 8.1 @math-11-nbf:sol:unit08
- an. ===== Question 7===== Given $\alpha$ and $\beta$ are acute angles with $\sin \alpha=\dfrac{12}{13}$ and $\tan \beta=\dfrac{4}{3}$ find \\ (i) $\sin(\alpha+\beta)$ (ii) $\cos(\alpha+\beta)$ (iii) $\tan(\alpha+\beta)$. ** Solution. ** Given: $\sin \alpha=\dfra