Question

If $cosB/sinA=n$ and $cosB/cosA=m$ Show that $(m^2+n^2)co{s}^{2}A=n^2$

Answer 1

$\frac{cosB}{sinA}=n$ $\frac{cosB}{cosA}=m$

We have to prove that, $(m^2+n^2)co{s}^{2}A=n^2$

So, $m^2+n^2=\frac{co{s}^{2}B}{si{n}^{2}A}+\frac{co{s}^{2}B}{co{s}^{2}A}$

$=co{s}^{2}B(\frac{1}{si{n}^{2}A}+\frac{1}{co{s}^{2}A})$

$=co{s}^{2}B\left(\frac{si{n}^{2}A+co{s}^{2}A}{si{n}^{2}A+co{s}^{2}A}\right)$

$=\frac{co{s}^{2}B}{si{n}^{2}Aco{s}^{2}A}$

$(m^2+n^2)co{s}^{2}A=\frac{co{s}^{2}B}{si{n}^{2}A}=n^2$

Hence proved