Question

If m tanA=tan mA , show that (sin^2 mA)÷(sin^2 A) = (m^2)÷{1+((m^2)-1) sin^2A}

Answer 1

$\textbf{Given:}$

$m\,tanA=tan\,mA$

$\textbf{To prove:}$

$\dfrac{sin^2mA}{sin^2A}=\dfrac{m^2}{1+(m^2-1)sin^2A}$

$\textbf{Solution:}$

$\text{Consider,}$

$m\,tanA=tan\,mA$

$\implies\,m=\dfrac{tan\,mA}{tanA}$

$\text{Now,}$

$\dfrac{m^2}{1+(m^2-1)sin^2A}$

$=\dfrac{\frac{tan^2mA}{tan^2A}}{1+(\frac{tan^2mA}{tan^2A}-1)sin^2A}$

$=\dfrac{tan^2mA}{tan^2A+(tan^2mA-tan^2A)sin^2A}$

$=\dfrac{tan^2mA}{tan^2A+tan^2mA\,sin^2A-tan^2Asin^2A}$

$=\dfrac{tan^2mA}{tan^2A-tan^2Asin^2A+tan^2mA\,sin^2A}$

$=\dfrac{tan^2mA}{tan^2A(1-sin^2A)+tan^2mA\,sin^2A}$

$=\dfrac{tan^2mA}{tan^2A\,cos^2A+tan^2mA\,sin^2A}$

$=\dfrac{tan^2mA}{(\frac{sin^2A}{cos^2A})\,cos^2A+tan^2mA\,sin^2A}$

$=\dfrac{tan^2mA}{sin^2A+tan^2mA\,sin^2A}$

$=\dfrac{tan^2mA}{sin^2A(1+tan^2mA)}$

$=\dfrac{tan^2mA}{sin^2A\,sec^2mA}$

$=\dfrac{\frac{sin^2mA}{cos^2mA}}{sin^2A(\frac{1}{cos^2mA})}$

$=\dfrac{sin^2mA}{sin^2A}$

$\implies\bf\dfrac{sin^2mA}{sin^2A}=\dfrac{m^2}{1+(m^2-1)sin^2A}$

Find more:

If l tan A + m sec A = n and l’tan A – m’sec A = n’, then show that (nl’ – ln’/ml’ + lm’)^2 = 1 + (nm’ – mn’/ lm’ + ml’) ^2

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