(0.5)/((+-sqrt(a^(2)-b^(2)),0)" won the line "bx cos x+ay sin x=ab?)

What is the product of the perpendicualars drawn from the point pm (sqrt(a^(2)-b^(2)),0) upon the line bx cos alpha + ay sin alpha =ab ?

Prove that the product of the length of the perpendiculars from the points (sqrt(a^2 -b^2) , 0) and (-sqrt(a^2 -b^2), 0) to the line x/a cos theta + y/a cos theta + y/b sin theta = 1 is

sin x+cos x=sqrt(2)sin5x

The area of the equilaterial triangle formed by the pair of lines (ax+by)^(2)-3(bx-ay)^(2)=0 and the line ax+by+c=0 is (c^(2))/(sqrt(3)(a^(2)+b^(2))) sq. units.

f(x)={((Sin(a+1)x+Sin2x)/(2x),,x lt 0),(b,,x=0),((sqrt(x-bx^3)-sqrtx)/(bx^(5/2) ), , xgt0):} , f(x) is a continous at x=0 then abs(a+b) is equal to

int(sin x cos x)/(sqrt(a^(2)sin^(2)x+b^(2)cos^(2)x))dx sin x

Let f: R to R be a function defined as f(x)={(,(sin (a+1)x+sin 2x)/(2x), if x lt 0),(, b, if x=0), (,(sqrt(x+bx^(3))-sqrtx)/(bx^(5/2)), if x gt 0):} If f is continuous at x=0 then the value of a+b is equal to :