An arithmetic progression is a sequence of numbers such that the difference of any two successive members is a constant
Let us consider following series
(1)1,5,9,13,17....
(2)1,2,3,4,5,...
(3)7,7,7,7.....
All these sets follow certain rules. In first set 5−1=9−5=13−9=17−13=4
In second set 2−1=3−2=4−3=1
and so on.
Here the difference between any successive members is a constant
Such series are called Arithmetic Progression

 cosec a -cot a =1/3

 cosec² a -cot²a =1

 cosec a +cot a =cosec² a -cot²a /cosec a -cot a

=1/1/3=3

Since ABC is right angled and angle C is 90°
therefore,
A+B=180°-C
A+B=180°-90°
A+B= 90°
Therefore,cos (A+B)= cos90° =0

let f(x)=kx²+5x+2

{tex}\begin{array}{l}\mathrm\alpha\;\;\mathrm{are}\;\mathrm\beta\;\mathrm{are}\;\mathrm{zer}o\;\mathrm{of}\;\mathrm f(\mathrm x)\\\mathrm{so}\;\alpha+\mathrm\beta=-\frac5{\mathrm k},\mathrm{αβ}=\frac2{\mathrm k}\end{array}{/tex}

{tex}\begin{array}{l}\frac1{\alpha^2}+\frac1{\beta^2}\\=\frac{\alpha^2+\beta^2}{\alpha^2\beta^2}=\frac{{\displaystyle\left(\alpha+\beta\right)^2}{\displaystyle-}{\displaystyle2}{\displaystyle\alpha}{\displaystyle\beta}}{\alpha^2\beta^2}\\=\frac{\displaystyle25/k^2-4/k}{\displaystyle\left({\displaystyle\frac2k}\right)^2}=\frac{\displaystyle\frac{25-4k}{k^2}}{\displaystyle\frac4{k^2}}=\frac{25-4k}4=\frac{17}4\\\\\end{array}{/tex}

{tex}\begin{array}{l}4\ast(25-4k)=17\ast4\\100-16k=68\\16k=100-68\\16k=32\\k=2\\\\\\\end{array}{/tex}

{tex}\alpha \ and\ \beta{/tex} are the roots of the polynomial, p(y) = 25p² - 15p + 2

a = 25,  b= -15,  c= 2
{tex} \alpha + \beta =-\frac{b}{a}= - \left( - \frac { 15 } {25 } \right) = \frac 35{/tex}
and {tex}\alpha \beta = \frac{c}{a} = \frac { 2 } { 25 }{/tex}
Now, {tex}\frac { 1 } { \alpha } + \frac { 1 } { \beta } = \frac { \alpha + \beta } { \alpha \beta } = \frac { 3 / 5 } { 2 /25 } = \frac { 15 } { 2 }{/tex}
and {tex}\frac { 1 } { \alpha } \times \frac { 1 } { \beta } = \frac { 1 } { \alpha \beta } = \frac { 1 } { 2 / 25 } = \frac{25}{2}{/tex}.
The equation of polynomial which has {tex}\frac{1}{\alpha} \ and \ \frac{1}{\beta}{/tex}as roots is {tex}y ^ { 2 } - \frac { 15 } {2 } y + \frac{25}{2} {/tex}

{tex}\frac { \cos 30 ^ { \circ } + \sin 60 ^ { \circ } } { 1 + \sin 30 ^ { \circ } + \cos 60 ^ { \circ } }{/tex}
{tex}= \frac { \frac { \sqrt { 3 } } { 2 } + \frac { \sqrt { 3 } } { 2 } } { 1 + \frac { 1 } { 2 } + \frac { 1 } { 2 } } = \frac { \sqrt { 3 } } { 2 }{/tex}
 {tex}\Rightarrow\cos 30 ^ { \circ } = \frac { \sqrt { 3 } } { 2 }{/tex}
Hence, {tex}\frac { \cos 30 ^ { \circ } + \sin 60 ^ { \circ } } { 1 + \sin 30 ^ { \circ } + \cos 60 ^ { \circ } } = \cos 30 ^ { \circ }{/tex}

tanA=cotB

tanA=1/tanB

tanAtanB=1

tan(A+B)=tanA+tanB/1-tanATab

=tanA+tanB/1-1

=tanA+tanB/0

tan(A+B)={tex}\infty{/tex}

or tan(A+B)=tan90

hence A+B=90

It is given that {tex} \alpha{/tex} and {tex} \beta{/tex} are the zeros of the quadratic polynomial {tex}f(x)=ax^2+bx+c{/tex} 
{tex} \therefore \quad \alpha + \beta = - \frac { b } { a } \text { and } \alpha \beta = \frac { c } { a }{/tex}
Now,

1. {tex}\alpha^3 + \beta ^3{/tex}
   = {tex}(\alpha + \beta )(\alpha^2 -\alpha\beta + \beta ^2){/tex}
   = {tex}(\alpha + \beta)( \alpha^2+\beta^2-\alpha\beta+2\alpha\beta-2\alpha\beta){/tex}
   = {tex} (\alpha + \beta)[(\alpha + \beta)^2-3\alpha \beta]{/tex}
   = {tex}(\alpha + \beta)^3-3\alpha \beta(\alpha + \beta){/tex}
   = {tex} \frac { - b ^ { 3 } +3abc} { a ^ { 3 } }{/tex}
   {tex}=\frac{-b^3+3abc}{a^3}{/tex} 

2. {tex} \frac { 1 } { \alpha ^ { 3 } } + \frac { 1 } { \beta ^ { 3 } } = \frac { \alpha ^ { 3 } + \beta ^ { 3 } } { ( \alpha \beta ) ^ { 3 } }{/tex}

   {tex}=\frac { (\alpha + \beta)( \alpha^2+\beta^2-\alpha\beta) } { ( \alpha \beta ) ^ { 3 } }{/tex}

   {tex}=\frac { (\alpha + \beta)( \alpha^2+\beta^2-\alpha\beta+2\alpha\beta-2\alpha\beta) } { ( \alpha \beta ) ^ { 3 } }{/tex}

   {tex}=\frac { (\alpha + \beta)[(\alpha + \beta)^2-3\alpha \beta] } { ( \alpha \beta ) ^ { 3 } }{/tex}

   {tex}=\frac { [(\alpha + \beta)^3-3\alpha \beta(\alpha + \beta)] } { ( \alpha \beta ) ^ { 3 } }{/tex}

   {tex} = \frac { \frac { - b ^ { 3 } +3abc} { a ^ { 3 } } } { \left( \frac { c } { a } \right) ^ { 3 } } {/tex}
   {tex}=\frac{-b^3+3abc}{a^3}\;\times\;\frac{a^3}{c^3}{/tex}
   {tex}= \frac { 3 a b c - b ^ { 3 } } { c ^ { 3 } }{/tex}

Let us suppose that one man alone can finish the work in x days and one boy alone can finish it in y days. Then,
one man's one day's work {tex}= \frac { 1 } { x }{/tex}
One boy's one day's work {tex}= \frac { 1 } { y }{/tex}
{tex}\therefore{/tex} Eight men's one day's work = {tex}\frac { 8 } { x }{/tex}
{tex}12\ boy's{/tex} one day's work = {tex}\frac { 12 } { y }{/tex}
According to question it is given that  {tex}8\ men{/tex} and {tex}12\ boys{/tex} can finish the work in {tex}10\ days{/tex}
{tex}10 \left( \frac { 8 } { x } + \frac { 12 } { y } \right) = 1 \Rightarrow \frac { 80 } { x } + \frac { 120 } { y } = 1{/tex} .................(i)
Again, {tex}6\ men{/tex} and {tex}8\ boys{/tex} can finish the work in {tex}14\ days{/tex}.
{tex}\therefore \quad 14 \left( \frac { 6 } { x } + \frac { 8 } { y } \right) = 1 \Rightarrow \frac { 84 } { x } + \frac { 112 } { y } = 1{/tex} ...........(ii)
Putting {tex}\frac { 1 } { x } = u{/tex} and {tex}\frac { 1 } { y } = v{/tex} in equations (i) and (ii), we get
{tex}80u + 120u - 1 = 0{/tex}
{tex}84u + 112v - 1 = 0{/tex}
By using cross-multiplication, 
{tex}\Rightarrow \frac { u } { - 120 + 112 } = \frac { - v } { - 80 + 84 } = \frac { 1 } { 80 \times 112 - 120 \times 84 }{/tex}
{tex}\Rightarrow \quad \frac { u } { - 8 } = \frac { v } { - 4 } = \frac { 1 } { - 1120 }{/tex}
{tex}\Rightarrow \quad u = \frac { - 8 } { - 1120 } = \frac { 1 } { 140 } \text { and } v = \frac { - 4 } { - 1120 } = \frac { 1 } { 280 }{/tex}
{tex}u = \frac { 1 } { 140 } \Rightarrow \frac { 1 } { x } = \frac { 1 } { 140 } \Rightarrow x = 140{/tex}
{tex}v = \frac { 1 } { 280 } \Rightarrow \frac { 1 } { y } = \frac { 1 } { 280 } \Rightarrow y = 280{/tex}
One man alone can finish the work in {tex}140\ days{/tex} and one boy alone can finish the work in {tex}280\ days{/tex}.

The given system of equations may be written as
{tex}9x -10y + 12 = 0{/tex} ...(i)
{tex}2x + 3y - 13 = 0{/tex}.... (ii)
From (ii), we get {tex} y = \frac { 13 - 2 x } { 3 }{/tex}
Substituting {tex} y = \frac { 13 - 2 x } { 3 }{/tex} in (i), we get
{tex} 9 x - \frac { 10 ( 13 - 2 x ) } { 3 } + 12 = 0{/tex}
{tex} \Rightarrow{/tex} {tex}27x - 10(13 - 2x) + 36 = 0{/tex}
{tex} \Rightarrow{/tex} {tex}27x -130 + 20x + 36 = 0{/tex}
{tex} \Rightarrow{/tex} {tex}47x - 94 = 0{/tex}

{tex} \Rightarrow{/tex} {tex}47x = 94{/tex}

{tex} \Rightarrow x = \frac { 94 } { 47 } = 2{/tex}
Substituting {tex}x = 2{/tex} in (i), we get
9 {tex} \times{/tex} 2 - 10y + 12 = 0 

{tex} \Rightarrow 10 y = 30 {/tex}

{tex}\Rightarrow y = \frac { 30 } { 10 } = 3{/tex}
Hence, x = 2 and y = 3 is the required solution.

Euclid division lemma:-
a=bq+r
324 = 96(3) + 36
96 = 36(2) + 24
36 = 24(1) + 12
24 = 12(2) + 0

hcf of 324 and 96 is 12.

Now find the hcf of 12 and 56,
56 = 12(4) + 8
12 = 8(1) + 4
8 = 4(2)+0
hcf of 12 and 56 is 4.

Therefore, hcf of 56.96 and 324 is 4