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प्रश्न
If y = sin (sin x), prove that \[\frac{d^2 y}{d x^2} + \tan x \frac{dy}{dx} + y \cos^2 x = 0 .\]
उत्तर
\[y = \sin(\sin x)\]
\[\frac{dy}{dx} = \cos(\sin x) . \cos x\]
\[\frac{d^2 y}{d x^2} = \cos(\sin x) . ( - \sin x) + \cos x . { - \sin(\sin x)} . \cos x = - \sin x . \cos(\sin x) - y \cos^2 x\]
\[\text { Now,} \]
\[\frac{d^2 y}{d x^2} + \tan x . \frac{dy}{dx} + y \cos^2 x\]
\[ = - \sin x . \cos(\sin x) - y \cos^2 x + \frac{\sin x}{\cos x} . \cos(\sin x) . \cos x + y \cos^2 x\]
\[ = - \sin x . \cos(\sin x) - y \cos^2 x + \sin x . \cos(\sin x) + y \cos^2 x\]
\[ = 0 .\]
Hence proved.
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