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प्रश्न
Prove that:
\[\frac{\cos^2 A}{1 - \tan A} + \frac{\sin^3 A}{\sin A - \cos A} = 1 + \sin A \cos A\]
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उत्तर
Solving L.H.S. of the equation:
\[\Rightarrow \frac{\cos^2 A}{1 - \tan A} + \frac{\sin^3 A}{\sin A - \cos A}\]
\[\Rightarrow \frac{\cos^2 A}{1 - \frac{\sin A}{\cos A}} + \frac{\sin^3 A}{\sin A - \cos A}\]
\[\Rightarrow \frac{\cos^2 A}{\frac{\cos A - \sin A}{\cos A}} + \frac{\sin^3 A}{\sin A - \cos A}\]
\[\Rightarrow \frac{\cos^3 A}{\cos A - \sin A} + \frac{\sin^3 A}{\sin A - \cos A}\]
\[\Rightarrow -\frac{\cos^3 A}{\sin A - \cos A} + \frac{\sin^3 A}{\sin A - \cos A}\]
\[\Rightarrow \frac{\sin^3 A - \cos^3 A}{\sin A - \cos A}\]
By formula,
\[a^3 - b^3 = (a - b)(a^2 + ab + b^2)\]
\[\Rightarrow \frac{(\sin A - \cos A)(\sin^2 A + \sin A \cos A + \cos^2 A)}{\sin A - \cos A}\]
\[\Rightarrow \sin^2 A + \cos^2 A + \sin A \cos A\]
By formula,
\[\sin^2 A + \cos^2 A = 1.\]
\[\Rightarrow 1 + \sin A \cos A.\]
Since, L.H.S. = R.H.S.
