Question

Find the distance between (acosa,0)(0,asina)

Answer 1

distance between two point A(x,y) & B(p,q)

is given by

=square root of [(x-p)²+(y-q)²]

=√[(x-p)²+(y-q)²]

Sony this expression

distance between (acosA+bsinA,0) & (0,asinA-bcosA)

is given by

=square root [( acosA+bsinA-0)²+(0-(asinA-bcosA))²]

=square root[( acosA+bsinA)²+(-asinA+bcosA)²]

=square root[( acosA+bsinA)²+(bcosA-asinA)²]

=square root[( a²cos²A+b²sin²A+2abcosAsinA)+(b²cos²A+a²sin²A-2abcosAsinA)]

by grouping terms

=square root[( a²cos²A+a²sin²A+2abcosAsinA)+(b²cos²A+b²sin²A-2abcosAsinA)]

=square root[( a²(cos²A+sin²A)+b²(cos²A+sin²A)]

=square root[( a²*1+b²*1]

=√(a²+b²)

Answer 2

Answer:

D = √(x1-x2)²+(y1-y2)²

D = √(acosa-0)² + (asina²-0)

( since bracket part is in square , we can take any value before , it doesn't make any difference)

D = √(a²cos²a + a² sin²a)

D = √a²(cos²a+sin²a)

D = √a²

D = a (Ans.)

$\color{red}{Mark\: as\: Brainlist}$