trigonometry

mun · **Joined:** Fri, 28 Dec 2007 12:01:53 UTC **Posts:** 1261

if $ah sin\theta +bkcos\theta=a^2-b^2$ .find total number of solution.and show that sum of roots are odd multiple of $\frac{\pi}{2}$ .here b,h,a,k

are real numbers

outermeasure · **Posted:** Tue, 31 Jan 2012 17:18:10 UTC

mun wrote:

if $ah sin\theta +bkcos\theta=a^2-b^2$ .find total number of solution.and show that sum of roots are odd multiple of $\frac{\pi}{2}$ .here b,h,a,k

are real numbers

Huh? There are either infinitely many, or no solutions. So the sum of roots, when defined, is zero and hence not an odd multiple of $\frac{\pi}{2}$ .

mun · **Joined:** Fri, 28 Dec 2007 12:01:53 UTC **Posts:** 1261

outermeasure wrote:

mun wrote:

if $ah sin\theta +bkcos\theta=a^2-b^2$ .find total number of solution.and show that sum of roots are odd multiple of $\frac{\pi}{2}$ .here b,h,a,k

are real numbers

Huh? There are either infinitely many, or no solutions. So the sum of roots, when defined, is zero and hence not an odd multiple of $\frac{\pi}{2}$ .

sir what happen if i replace + sign by - .
$ah sin\theta -bkcos\theta=a^2-b^2$

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