Given tanA = m tanB
\(\cfrac { sinA }{ cosA } =m\cfrac { sinB }{ cosB } \) 
\(\cfrac { sinAcosB }{ cosAsinB } =m\) 
Applying componendo and dividendo rule, we get
\(\cfrac { sinAcosB+cosAsinB }{ sinAcosB-cosAsinB } =\cfrac { m+1 }{ m-1 } \) 
\(\cfrac { sin(A+B) }{ sin(A-B) } =\cfrac { m+1 }{ m-1 } \)
which completes the proof.

In the word MATHEMATICS
There are 11 letters
Here M occurs twice
T occurs twice
A occurs twice and
the rest are all different. Therefore number of permutations
= \(\cfrac { 11! }{ 2!2!21 } \)

Here \(a=2,b=2,h=\cfrac { 5 }{ 2 } ,g=\cfrac { 15 }{ 2 } ,f=9,c=k\) 
The given line represents a pair of straight lines if,
\(abc+2fgh-{ af }^{ 2 }-{ bg }^{ 2 }-{ ch }^{ 2 }=0\)
\(i.e., \ 4k+\cfrac { 675 }{ 2 } -162\cfrac { 225 }{ 2 } -\cfrac { 25 }{ 4 } k=0\)
\( \Rightarrow 16k+1350-648-450-25k=0\)
\(\Rightarrow 9k=252\therefore k=28\)

\(A=\left[ \begin{matrix} 2 & 4 \\ -3 & 2 \end{matrix} \right] \)
\(|A|=\left[ \begin{matrix} 2 & 4 \\ -3 & 2 \end{matrix} \right] \)
=16 ≠ 0
Since A is a nonsingular matrix, A ^-1 exists
Now  adj \(A=\left[ \begin{matrix} 2 & -4 \\3 & 2 \end{matrix} \right] \)
\({ A }^{ -1 }=\frac { 1 }{ \left| A \right| } adjA\)
\(=\frac { 1 }{ 16 } \left[ \begin{matrix} 2 & -4 \\ 3 & 2 \end{matrix} \right] \)

n(S) = 52
(i) Probability of drawing 2 ace cards without replacement
\(=\frac { 4 }{ 52 } \times \frac { 3 }{ 51 } =\frac { 1 }{ 221 } \)
(ii) Probability of drawing 2 spades without replacement
\(=\frac { 13 }{ 52 } \times \frac { 12 }{ 51 } =\frac { 3 }{ 51 } =\frac { 1 }{ 17 } \)

\(\bar{X}\) = 6, \(\bar{Y}\) = 8, \(\sigma _x=5\)
\(\sigma_y=\frac{40}{3}, \quad r=\frac{8}{15} \)
\(b_{y x}=r \frac{\sigma_x}{\sigma_y}=\frac{8}{15}\left(\frac{40}{3 \times 5}\right)=\frac{64}{45}=1.422\)
Regression line of Y on X is
\(Y-\bar{Y}=b_{y x}(X-\bar{X}) \)
Y - 8 = 1.422(X - 6)
Y = 1.422 X - 8.532 + 8
Y = 1.422 X - 0.532 .
If  X = Rs 100,
Y = 142.2 - 0.532 = Rs 141.67

a = Rs. 10,000, i = \(\cfrac{10}{200}\) = 0.05, n = 5 x 2 = 10
P = \(\frac { a }{ i } \left[ \left( 1+i \right) ^{ n }-1 \right] \)
= \(\frac { 10,000 }{ 0.05 } \left[ \left( 1.05 \right) ^{ 10 }-1 \right] \)
= 200000 [1.6289 - 1]
= 200000 [0.6289]
= Rs. 125,780

Annual demand : R = 4000
Storage cost: C₁ = Rs. 50
Setup cost per production: C₃= Rs. 160
EOQ = \(\sqrt { \frac { 2R{ c }_{ 3 } }{ { c }_{ 1 } } } \) = \(\sqrt { \frac { 2\times 4000\times 160 }{ 50 } } =160\)
\(\therefore\) To minimize the production cost number of units produced per run is 160 units.

i) variables:
Let x₁, x₂ be the number of pills of size A and B respectively.
ii) Objective function:
The cost is to be minimized
∴ minimize Z = x₁ + x₂
iii) Constraints:
The data can be summarised as follows

\(2 x_1+x_2 \geq 12 \ [\because\) Chemical is found by users that it requires at 12 mgs of aspirin]
\(5 x_1+8 x_2 \geq 74 \ [\because\) Chemical is found by users that it requires at 74 mgs of Bicarbonate]
\(x_1+6 x_2 \geq 24 \ [\because\) Chemical is found by users that it requires at 24 mgs of codeine]
(iii) Non-negative restrictions:
Since the number of pills of size A and B are non negative, we have x₁ + x₂  ​\(\le \)​  0
Thus, we have the following linear programming model.
Minimize  \(z=x_1+x_2\)   
Subject to constraints 
\(\\ 2x_{ 1 }+{ x }_{ 2 }\ge 12\)
\(5x_{ 1 }+8{ x }_{ 2 }\ge 74\)
\(x_{ 1 }+6{ x }_{ 2 }\ge 24\)
\({ x }_{ 1 },{ x }_{ 2\ }\le \ 0\)

Let y = sin (x²)
⇒ \(\frac { dy }{ dx } = \cos (x^2).2x\)
= 2x.(cos(x²))