Suppose that the digits at units and tens place of the given number be x and y respectively.
Thus, the number is $10y + x.$
The product of the two digits of the number is 20.
Thus, we have $xy = 20$
After interchanging the digits, the number becomes $10x + y$
If 9 is added to the number, the digits interchange their places.
Thus, we have
$(10y + x) + 9 = 10x + y$
$\Rightarrow$ $10y + x + 9 = 10x+ y$
$\Rightarrow$ $10x + y - 10y - x = 9$
$9x - 9y = 9$
$\Rightarrow$$9(x - y) = 9$
$\Rightarrow x - y = \frac{9}{9}$
$\Rightarrow$ $x - y = 1$
So, we have the systems of equations
$xy = 20$ ....(i)
$x - y = 1$ ....(ii)
Here x and y are unknowns.
We have to solve the above systems of equations for x and y.
Substituting $x = 1 + y$ from the second equation to the first equation, we get $(1+ y) y = 20$
$\Rightarrow$ $y + y^2 = 20$
$\Rightarrow$ $y^2 + y - 20 = 0$
$\Rightarrow$ $y^2 + 5y - 4y - 20 = 0$
$\Rightarrow$ $y(y + 5) - 4(y + 5) = 0$
$\Rightarrow$ $(y + 5)(y - 4) = 0$
$\Rightarrow$ $y = -5\ or\ y = 4$
Substituting the value of y in the second equation, we have

Note that in the first pair of solution the values of x and y are both negative. But the digits of the number can't be negative. So, we must remove this pair.
Hence, the number is 10 $\times$ 4 + 5 = 45

For sin A,
By using identity $cosec ^ { 2 } A - \cot ^ { 2 } A = 1 \Rightarrow \cos e c ^ { 2 } A = 1 + \cot ^ { 2 } A$
$\Rightarrow \frac { 1 } { \sin ^ { 2 } A } = 1 + \cot ^ { 2 } A$
$\Rightarrow \sin A = \frac { 1 } { \sqrt { 1 + \cot ^ { 2 } A } }$
For secA, 
By using identity $\sec ^ { 2 } A - \tan ^ { 2 } A = 1 \Rightarrow \sec ^ { 2 } A = 1 + \tan ^ { 2 } A$
$\Rightarrow \sec ^ { 2 } A = 1 + \frac { 1 } { \cot ^ { 2 } A } = \frac { \cot ^ { 2 } A + 1 } { \cot ^ { 2 } A } \Rightarrow \sec ^ { 2 } A = \frac { 1 + \cot ^ { 2 } A } { \cot ^ { 2 } A }$
$\Rightarrow \sec A = \frac { \sqrt { 1 + \cot ^ { 2 } A } } { \cot A }$
For tanA,
$\tan A = \frac { 1 } { \cot A }$

Given that at the foot of a mountain the elevation of its summit is 45°; after ascending 1000 m towards the mountain up a slope of 30° inclination, the elevation is found to be 60°. We have to find the height of the mountain.
Let F be the foot and S be the summit of the mountain FOS. Then $\angle O F S = 45 ^ { \circ }$and therefore, $\angle O S F = 45 ^ { \circ }.$Consequently, OF = OS = h km (say). Let FP = 1000 m = 1 km be the slope so that $\angle O F P = 30 ^ { \circ }.$Draw PM $\perp$OF. join PS. It is given that $\angle M P S = 60 ^ { \circ }.$
In $\triangle F P L,$we have

$\sin 30 ^ { \circ } = \frac { P L } { P F }$
$\Rightarrow \quad P L = P F \sin 30 ^ { \circ } = \left( 1 + \frac { 1 } { 2 } \right) \mathrm { km } = \frac { 1 } { 2 } \mathrm { km }$
$\therefore \quad O M = P L = \frac { 1 } { 2 } \mathrm { km }$
$\Rightarrow \quad M S = O S - O M = \left( h - \frac { 1 } { 2 } \right) \mathrm { km }$ ...(i)
Also, $\cos 30 ^ { \circ } = \frac { F L } { P F }$
$\Rightarrow \quad F L = P F \cos 30 ^ { \circ } = \left( 1 \times \frac { \sqrt { 3 } } { 2 } \right) \mathrm { km } = \frac { \sqrt { 3 } } { 2 } \mathrm { km }$
Now, h = OS = OF = OL + LF
$\Rightarrow \quad h = O L + \frac { \sqrt { 3 } } { 2 }$
$\Rightarrow \quad O L = \left( h - \frac { \sqrt { 3 } } { 2 } \right) \mathrm { km }$
$\Rightarrow \quad P M = \left( h - \frac { \sqrt { 3 } } { 2 } \right) \mathrm { km }$ ...(ii)
In $\triangle S P M,$ we have
$\tan 60 ^ { \circ } = \frac { S M } { P M }$
$\Rightarrow$ SM = PM . tan60 ^° 
$\Rightarrow \quad \left( h - \frac { 1 } { 2 } \right) = \left( h - \frac { \sqrt { 3 } } { 2 } \right) \sqrt { 3 }$
$\Rightarrow \quad h - \frac { 1 } { 2 } = h \sqrt { 3 } - \frac { 3 } { 2 }$
$\Rightarrow \quad \sqrt { 3 } h - h = \frac { 3 } { 2 } - \frac { 1 } { 2 }$
$\Rightarrow \quad h ( \sqrt { 3 } - 1 ) = 1$
$\Rightarrow \quad h = \frac { 1 } { \sqrt { 3 } - 1 } = \frac { \sqrt { 3 } + 1 } { ( \sqrt { 3 } - 1 ) ( \sqrt { 3 } + 1 ) } = \frac { \sqrt { 3 } + 1 } { 2 } = \frac { 2.732 } { 2 } = 1.366 \mathrm { km }$
Hence, the height of the mountain is 1.366 km.

Let a be the first term and d the common difference of the given A.P.
$\therefore S_{p}=\frac{p}{2}$ [2a + (p - 1)d] = q 
$\Rightarrow$ 2a + (p - 1)d $=\frac{2 q}{p}$  ….(i)
And $S_{q}=\frac{q}{2}$ [2a + (q - 1)d] = p
$\Rightarrow$ 2a + (q - 1)d $=\frac{2 p}{q}$ ….(ii)
Subtracting eq. (ii) from eq. (i) we get
(p - q)d = $\frac{2 q}{p}-\frac{2 p}{q}$ $\Rightarrow$ (p - q)d $=\frac{2\left(q^{2}-p^{2}\right)}{p q}$
$\Rightarrow$ (p - q)d $=\frac{-2}{p q}$(p² - q²)
$\Rightarrow$ (p - q)d $=\frac{-2}{p q}$ (p + q)(p - q) $\Rightarrow d=\frac{-2}{p q}$ (p + q)
Substituting the value of d in eq. (i) we get
2a + (p - 1) $\left[\frac{-2(p+q)}{p q}\right]=\frac{2 q}{p}$
$\Rightarrow 2 a=\frac{2 q}{p}+\frac{2(p-1)(p+q)}{p q}$
$\Rightarrow a=\frac{q}{p}+\frac{(p-1)(p+q)}{p q}$
$a=\frac{q^{2}+p^{2}+p q-p-q}{p q}$
Now S_p+q $=\frac{p+q}{2}$ [2a + (p + q - 1)d
$=\frac{p+q}{2}\left[\frac{2 q^{2}+2 p^{2}+2 p q-2 q-2 q}{p q}+\frac{(p+q-1)[-2(p+q)}{p q}\right]$
$=\frac{p+q}{2}\left[\frac{2q^{2} + 2p^{2} + 2pq - 2p - 2q -2p^{2} -2 p q+2 p-2 p q-2 q^{2}+2 q}{p q}\right]$
$=\frac{p+q}{2}\left[\frac{-2 p q}{p q}\right]$ = -(p + q) hence proved.

7 ones + 8 tenth = 7 + 80 = 87
 

a³ + b³ = (a + b) (a²- ab + b²).

The Ancient Greek mathematician Archimedes of Syracuse (287-212 BC) is largely considered to be the first to calculate an accurate estimation of the value of pi.