Question

heyaaa,,,

please Solve this question.....

If a line theta + b cos theta = c, then prove that a cos theta - b sin theta = A square + B square - C square. under root.

Answer 1

hey see the above pic
thanq

Answer 2

$a sin(x) + bcos(x) = c$
$\frac{asin(x) + bcos(x)}{\sqrt{a^2+b^2}} = \frac{c}{\sqrt{a^2+b^2}}$
let 
[tex]cos(t) = \frac{a}{ \sqrt{a^2+b^2} } [/tex]
then
$sin(t) = \frac{b}{ \sqrt{a^2+b^2} }$
so
$cos(t)sin(x)+sin(t)cos(x) = sin(t+x) = \frac{c}{ \sqrt{a^2+b^2} }$
now
[tex]a cos(x) - bsin(x) = \sqrt{a^2+b^2}(cos(t)cos(x)-sin(t)sin(x)) = \\ \sqrt{a^2+b^2} (cos(t+x)) = \sqrt{a^2+b^2} \frac{( \sqrt{a^2+b^2-c^2} )}{ \sqrt{a^2+b^2}} = \sqrt{a^2+b^2-c^2} [/tex]