If alpha = frac{sin^3 x}{cos^2 x},beta = frac | vedclass

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(B) Given,$\alpha = \frac{\sin^3 x}{\cos^2 x}$ and $\beta = \frac{\cos^3 x}{\sin^2 x}$.$\alpha \sin x + \beta \cos x + 3 = \frac{\sin^4 x}{\cos^2 x} +

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$\frac{4}{(k^2-1)^2}$

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(B) Given,$\alpha = \frac{\sin^3 x}{\cos^2 x}$ and $\beta = \frac{\cos^3 x}{\sin^2 x}$.$\alpha \sin x + \beta \cos x + 3 = \frac{\sin^4 x}{\cos^2 x} + \frac{\cos^4 x}{\sin^2 x} + 3$.$= \frac{\sin^6 x + \cos^6 x + 3 \sin^2 x \cos^2 x}{\sin^2 x \cos^2 x}$.Using the identity $a^3 + b^3 = (a+b)(a^2 - ab + b^2)$,where $a = \sin^2 x$ and $b = \cos^2 x$:$= \frac{(\sin^2 x + \cos^2 x)(\sin^4 x - \sin^2 x \cos^2 x + \cos^4 x) + 3 \sin^2 x \cos^2 x}{\sin^2 x \cos^2 x}$.$= \frac{(\sin^2 x + \cos^2 x)^2 - 3 \sin^2 x \cos^2 x + 3 \sin^2 x \cos^2 x}{\sin^2 x \cos^2 x} = \frac{1}{\sin^2 x \cos^2 x}$.Given $\sin x + \cos x = k$,squaring both sides gives $1 + 2 \sin x \cos x = k^2$,so $\sin x \cos x = \frac{k^2 - 1}{2}$.Thus,$\frac{1}{\sin^2 x \cos^2 x} = \frac{1}{(\frac{k^2 - 1}{2})^2} = \frac{4}{(k^2 - 1)^2}$.

If $\alpha = \frac{\sin^3 x}{\cos^2 x}$,$\beta = \frac{\cos^3 x}{\sin^2 x}$ and $\sin x + \cos x = k$,then $\alpha \sin x + \beta \cos x + 3 = $

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