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प्रश्न
If \[T_n = \sin^n x + \cos^n x\], prove that \[\frac{T_3 - T_5}{T_1} = \frac{T_5 - T_7}{T_3}\]
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उत्तर
LHS:\[\frac{T_3 - T_5}{T_1} = \frac{\left( \sin^3 x + \cos^3 x \right) - \left( \sin^5 x + \cos^5 x \right)}{\sin x + \cos x}\]
\[ = \frac{\sin^3 x - \sin^5 x + \cos^3 x - \cos^5 x}{\sin x + \cos x}\]
\[ = \frac{\sin^3 x\left( 1 - \sin^2 x \right) + \cos^3 x\left( 1 - \cos^2 x \right)}{\sin x + \cos x}\]
\[ = \frac{\sin^3 x . \cos^2 x + c {os}^3 x . \sin^2 x}{\sin x + \cos x}\]
\[ = \frac{\sin^2 x . \cos^2 x\left( \sin x + cos x \right)}{\sin x + \cos x}\]
\[ = \sin^2 x . \cos^2 x\]
RHS: \[\frac{T_5 - T_7}{T_3}\]
\[ = \frac{\left( \sin^5 x + \cos^5 x \right) - \left( \sin^7 x + \cos^7 x \right)}{\sin^3 x + \cos^3 x}\]
\[ = \frac{\sin^5 x - si n^7 x + \cos^5 x - \cos^7 x}{\sin^3 x + \cos^3 x}\]
\[ = \frac{\sin^5 x\left( 1 - \sin^2 x \right) + \cos^5 x\left( 1 - \cos^2 x \right)}{\sin^3 x + \cos^3 x}\]
\[ = \frac{\sin^5 x \cos^2 x + \cos^5 x \sin^2 x}{\sin^3 x + \cos^3 x}\]
\[ = \sin^2 x . \cos^2 x\]
LHS = RHS
Hence proved.
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