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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 5, Exercise 10.1 @fsc-part1-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 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: $A=\begin{bmatrix}x & y & z\end{bmatrix}$, $B=\begin{bma
- Question 4 Exercise 8.2 @math-11-nbf:sol:unit08
- \theta$ 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
- Exercise 6.1 @matric:9th_science
- d info 60%> We have created this page and it will be updated to add new solutions occasionally. Please... 3-2x^2)$, $54(27x^4-x)$ **Solution:**\\ (i) $\begin{align} x^2+5x+6&=x^2+3x+2x+6,\\ &=x(x+3)+2(x+3)\\ &=(x+3)(x+2) \end{align}$ $\begin{align} x^2-4x-12&=x^2-6x+2x-12,\\ &=x(x-6)+2(x... &=(x-6)(x+2) \end{align}$ H.C.F= $x+2$ (ii) $\begin{align} x^3-27 &=x^3-3^3,\\ &=(x-3)(x^2+3x+9)\e
- 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*}
- Question11 and 12, Exercise 10.1 @fsc-part1-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
- 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 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
- How to prepare admission test (A short guide) @papers:old_admission_test_of_assms_for_ph.d._mathematics
- Syllabus & Paper Pattern for General Mathematics (Split Program) @bsc:paper_pattern:punjab_university
- 1st SIBAU-NU International Workshop on Matrix Analysis and Linear Algebra (15-17 October 2021) @events
- Recent Advances in Mathematical Methods, Models & Applications, LSC Lahore, Pakistan (April 13-14, 2019) @conferences
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- International Conference on Computing and Mathematical Sciences, IBA Sukkur (February 25-26, 2017) @conferences
- 22nd International Pure Mathematics Conference on Algebra, Analysis and Geometry (23 to 25 August 2021) @events
- Chapter 10: Viewer @bsc:notes_of_mathematical_method:ch10_higher_order_linear_differential_equations
- View Online: BSc Mathematics (Old Papers) @papers:old_papers_for_bsc_mathematics:sargodha_university