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

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(B) આપેલ છે કે,$\alpha = \frac{\sin^3 x}{\cos^2 x}$ અને $\beta = \frac{\cos^3 x}{\sin^2 x}$.$\alpha \sin x + \beta \cos x + 3 = \frac{\sin^4 x}{\cos^2

Vedclass · Accepted Answer

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

Vedclass · Answer

(B) આપેલ છે કે,$\alpha = \frac{\sin^3 x}{\cos^2 x}$ અને $\beta = \frac{\cos^3 x}{\sin^2 x}$.$\alpha \sin x + \beta \cos x + 3 = \frac{\sin^4 x}{\cos^2 x} + \frac{\cos^4 x}{\sin^2 x} + 3$.$= \frac{\sin^6 x + \cos^6 x + 3 \sin^2 x \cos^2 x}{\sin^2 x \cos^2 x}$.નિત્યસમ $a^3 + b^3 = (a+b)(a^2 - ab + b^2)$ નો ઉપયોગ કરતા,જ્યાં $a = \sin^2 x$ અને $b = \cos^2 x$:$= \frac{(\sin^2 x + \cos^2 x)(\sin^4 x - \sin^2 x \cos^2 x + \cos^4 x) + 3 \sin^2 x \cos^2 x}{\sin^2 x \cos^2 x}$.$= \frac{(\sin^2 x + \cos^2 x)^2 - 3 \sin^2 x \cos^2 x + 3 \sin^2 x \cos^2 x}{\sin^2 x \cos^2 x} = \frac{1}{\sin^2 x \cos^2 x}$.આપેલ છે કે $\sin x + \cos x = k$,બંને બાજુ વર્ગ કરતા $1 + 2 \sin x \cos x = k^2$,તેથી $\sin x \cos x = \frac{k^2 - 1}{2}$.આમ,$\frac{1}{\sin^2 x \cos^2 x} = \frac{1}{(\frac{k^2 - 1}{2})^2} = \frac{4}{(k^2 - 1)^2}$.

જો $\alpha = \frac{\sin^3 x}{\cos^2 x}$,$\beta = \frac{\cos^3 x}{\sin^2 x}$ અને $\sin x + \cos x = k$ હોય,તો $\alpha \sin x + \beta \cos x + 3 = $

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