cosec A + sec A = cosec B + sec B
cosec A - cosec B  = sec B - sec A
\(\frac{1}{\sin A}-\frac{1}{\sin B}=\frac{1}{\cos B}-\frac{1}{\cos A}\)
\(\frac{\sin B-\sin A}{\sin A \sin B}=\frac{\cos A-\cos B}{\cos B \cos A}\)
\(\frac{\sin B-\sin A}{\cos A-\cos B}=\frac{\sin A \sin B}{\cos A \cdot \cos B}\)
\(\frac{2 \sin \frac{B-A}{2} \cos \frac{B+A}{2}}{-2 \sin \frac{A-B}{2} \sin \frac{A+B}{2}}=\tan A \tan B\)
\(\cot \left(\frac{A+B}{2}\right)=\tan A \tan B\)

Step-1 : Let P (n) denote the statement a - b is divisible by a - b for all \(n\in N\)
∴ a¹ - b¹ = a - b is divisible by a - b
Put n = 1
P(1) : a - b is divisible by a - b
∴ P(1) is true.
Step-2:
Let us assume that P(k) is true.
∴ a^k - b^k is divisible by a - b
⇒ a^k - b^k = C(a - b) where C is a constant     ...(1)
Step-3:
To prove that P(k+1) is true
P(k + 1) = a^k+1 - b^k+1
= a^k+1 - a^kb + a^kb - b^k+1  [Adding and subtracting ak b]
= a^k (a - b) + b (a^k - b^k)
= a^k (a - b) + b.c. (a - b) [using (1)]
= (a - b) (a^k + bc) which is divisible by a - b.
∴ P (k + 1) is true whenever P(k) is true
∴ P(n) is true for all \(n\in N\).

Given f(2) = 0
= 2 + 2
\(=4 \neq 0\)
\(\therefore \lim _{x \rightarrow 2} f(x) \neq f(2)\)
The function is not continuous at x = 2

Given equations are
x - y + 2z = 3, 2x + z = 1 and 3x + 2y +z = 4.
The given equations can be written in matrix form as 
\(\begin{bmatrix} 1&-1&2\\2&0&1\\3&2&1 \end{bmatrix}\begin{bmatrix} x\\y\\z \end{bmatrix}=\begin{bmatrix} 3\\1\\4 \end{bmatrix}\)
\(A X=B \Rightarrow X=A^{-1} B\)
\(\text {Where } A=\left(\begin{array}{ccc} 1 & -1 & 2 \\ 2 & 0 & 1 \\ 3 & 2 & 1 \end{array}\right), X=\left(\begin{array}{l} x \\ y \\ z \end{array}\right)\)
\(B=\left(\begin{array}{l} 3 \\ 1 \\ 4 \end{array}\right)\)
\(|A|=1(0-2)+1(2-3)+2(4-0)\)
\(=-2-1+8=5 \neq 0\)
\(\therefore A^{-1} \text { exists }\)
\(\text {Co-factor matrix }=\left(\begin{array}{ccc} -2 & 1 & 4 \\ 5 & -5 & -5 \\ -1 & 3 & 2 \end{array}\right)\)
\(\mathrm{A}^{-1}=\frac{1}{|A|} \operatorname{adj} \mathrm{A}=\frac{1}{5}\left(\begin{array}{ccc} -2 & 5 & -1 \\ 1 & -5 & 3 \\ 4 & -5 & 2 \end{array}\right)\)
\(\mathrm{X}=\mathrm{A}^{-1} \mathrm{~B}\)
\(={{1}\over{5}}\begin{bmatrix} -2&5&-1\\1&-5&3\\4&-5&2 \end{bmatrix}\begin{bmatrix} 3\\1\\4 \end{bmatrix}={{1}\over{5}}\begin{bmatrix} -6+5-4\\3-5+12\\12-5+8 \end{bmatrix}={{1}\over{5}}\begin{bmatrix} -5\\10\\15 \end{bmatrix}=\begin{bmatrix} -1\\2\\3 \end{bmatrix}\)
\(\left(\begin{array}{l} x \\ y \\ z \end{array}\right)=\left(\begin{array}{c} -1 \\ 2 \\ 3 \end{array}\right)\)
\(x=-1, y=2, z=3\)

\(A=\left(\begin{array}{ccc} 3 & -1 & 1 \\ -15 & 6 & -5 \\ 5 & -2 & 2 \end{array}\right)\)
\(|A|=3(12-10)+1(-30+25)+1(30-30)\)
\(=6-5=1 \neq 0\)
\(\therefore A^{-1} \text { exists }\)
\(\text {Co-factor matrix }=\left(\begin{array}{ccc} 2 & 5 & 0 \\ 0 & 1 & 1 \\ -1 & 0 & 3 \end{array}\right)\)
\(A^{-1}=\frac{1}{|A|} \operatorname{adj} A=\left(\begin{array}{ccc} 2 & 0 & -1 \\ 5 & 1 & 0 \\ 0 & 1 & 3 \end{array}\right)\)

\(\frac{x+2}{(x-1)(x+3)^2}=\frac{A}{(x-1)}+\frac{B}{(x+3)}+\frac{C}{(x+3)^2}\)
\(=\frac{A(x+3)^2+B(x-1)(x+3)+C(x-1)}{(x-1)(x+3)^2}\)
\(\Rightarrow\)  x + 2 = A ( x + 3 )² + B ( x - 1 ) ( x + 3 ) + C( x - 1 ) ....(1)
Putting x = 1 in (1) we get,
\(3=A(4)^2 \Rightarrow A=\frac{3}{16}\)
Puttingx = -3 in (1)we get
\(-1=C(-3-1)\)
\(C=\frac{1}{4}\)
Equate co-efficient of x² on both sides of (1)
\(0 =A+B \Rightarrow B=-A=\frac{-3}{16} \)
\(\frac{x+2}{(x-1)(x+3)^2} =\frac{3}{16(x-1)}-\frac{3}{16(x+3)}+\frac{1}{4(x+3)^2}\)

Let \(y=f\left( x \right) ={ x }^{ 2 }-5\)
Choose suitable values for x and determine y.
Thus we get the following table.

Plot the points (-3, 4), (-2, -1), (-1, -4), (0, -5), ( 1, -4), (2, -1), (3, 4) and draw a smooth curve.
The graph is as shown in the figure

A_ij = (–1)^i+jM_ij
A₁₁ = (–1)¹⁺¹M₁₁ = \(\left| \begin{matrix} 1 & 0 \\ 1 & 0 \end{matrix} \right| \) = 0
A₁₂ = (–1)¹⁺²M₁₂ = \(-\left| \begin{matrix} 0 & 0 \\ -4 & 0 \end{matrix} \right| \) = 0
A₁₃ = (–1)¹⁺³M₁₃ = \(\left| \begin{matrix} 0 & 1 \\ -4 & 1 \end{matrix} \right| \)= 0 - (-4) = 4
A₂₁ = (–1)²⁺¹M₂₁ = \(-\left| \begin{matrix} -2 & -3 \\ 1 & 0 \end{matrix} \right| \)= -(0 - (-3)) = -3
A₂₂ = (–1)²⁺²M₂₂ = \(\left| \begin{matrix} 1 & -3 \\ -4 & 0 \end{matrix} \right| \) = 0 - 12 = -12
A₂₃ = (–1)²⁺³M₂₃ = \(-\left| \begin{matrix} 1 & -2 \\ -4 & 1 \end{matrix} \right| \) = -(1-8) = 7
A₃₁ = (–1)³⁺¹M₃₁ = \(\left| \begin{matrix} -2 & -3 \\ 1 & 0 \end{matrix} \right| \) = 0 - (-3) = 3
A₃₂ = (–1)³⁺²M₃₂ = \(-\left| \begin{matrix} 1 & -3 \\ 0 & 0 \end{matrix} \right| \) = 0 - 0 = 0
A₃₃ = (–1)³⁺³M₃₃ = \(\left| \begin{matrix} 1 & -2 \\ 0 & 1 \end{matrix} \right| \) = 1 - 0 = 1
\(\left[ { A }_{ ij } \right] =\left[ \begin{matrix} 0 & 0 & 4 \\ -3 & -12 & 7 \\ 3 & 0 & 1 \end{matrix} \right] \)
Adj A = [A_ij]^T
\(=\left[ \begin{matrix} 0 & -3 & 3 \\ 0 & -12 & 0 \\ 4 & 7 & 1 \end{matrix} \right] \)

Y = -x² + 10 x - 15
\(\Rightarrow\) x² - 10x = - y - 15
\(\Rightarrow\) (x- 5)² = -y - 15 + 25
\(\Rightarrow\) (x- 5)² = -y + 10 
\(\Rightarrow\) (x - 5)² = (-y - 10)
The best time to end the project is when x = 5 months.

A lies in I quadrant and B lies in the IlI quadrant

\(\cos A=\sqrt{1-\sin ^2 A}\)
\(=\sqrt{1-\frac{9}{25}}=\sqrt{\frac{16}{25}}=\frac{4}{5}\)
\(\cos B=-\frac{12}{13}\)
\(\sin B=-\sqrt{1-\cos ^2 B}\)
\(=-\sqrt{1-\frac{144}{169}}=-\sqrt{\frac{25}{169}}=\frac{-5}{13}\)
cos (A + B)
= cos A cos B - sin A sin B
= \(\left( \frac { 4 }{ 5 } \right) \left( -\frac { 12 }{ 13 } \right) -\left( \frac { 3 }{ 5 } \right) \left( \frac { -5 }{ 13 } \right) \)
\(=\frac { -33 }{ 65 } \)