Question

Hi frndz. Solve this.

Answer 1

$\textsf{Given :}$

$sin\alpha=msin\beta$

$\implies sin^2\alpha=m^2sin^2\beta$

$\implies \frac{m^2}{sin^2\alpha}=\frac{1}{sin^2\beta}$

Using the identity : $sinA=\frac{1}{cosecA}$

$\implies m^2cosec^2\alpha=cosec^2\beta$ ...........( 1 )

$\textsf{Also it is given that : }$

$tan\alpha=ntan\beta$

$\implies tan^2\alpha=n^2tan^2\beta$

$\implies \frac{n^2}{tan^2\alpha}=\frac{1}{cot^2\beta}$

Using the identity : $tanA=\frac{1}{cotA}$

$\implies n^2cot^2\alpha}=cot^2\beta$ ...........( 2 )

Subtracting ( 2 ) from ( 1 ) we get :

$cosec^2\beta-cot^2\beta=m^2cosec^2\alpha-n^2cot^2\beta$

$\implies 1 = m^2cosec^2\alpha-n^2cot^2\alpha$

Use these identities : $cosecA=\frac{1}{sinA}$

                                    $cotA=\frac{cosA}{sinA}$

$\implies \frac{m^2}{sin^2\alpha}-\frac{n^2cos^2\alpha}{sin^2\alpha}=1$

$\implies \frac{1}{sin^2\alpha}(m^2-n^2cos^2\alpha)=1$

$\implies m^2-n^2cos^2\alpha=sin^2\alpha$

$\implies m^2-n^2cos^2\alpha=1-cos^2\alpha$

$\implies cos^2\alpha-n^2\alpha=1-m^2$

$\implies cos^2\alpha(1-n^2)=1-m^2$

$\implies cos^2\alpha=\frac{1-m^2}{1-n^2}$

$\implies cos^2\alpha=\frac{(-1)(m^2-1)}{(-1)(n^2-1)}$

$\implies \boxed{cos^2\alpha=\frac{m^2-1}{n^2-1}}$[P.R.O.V.E.D]

Hope I helped u :-)

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