Question

a sinx + b cosx= a cosecx + b secx, then prove that the value of each side is (a^2/3 - b^2/3)√(a^2/3 + b^2/3)​

Answer 1

$\tt \underline{\underline{\red{Given:-}}}$

• a sinx + b cosx = a cosecx + b secx

$\tt \underline{\underline{\green{To \: Prove:-}}}$

$\bullet \tt \: (a^{2/3} - b^{2/3}) \sqrt{(a^{2/3 }+ b^{2/3})}$

$\tt \underline{\underline{\orange{Proof:-}}}$

$\tt \hookrightarrow \: a \: sinx + b \: cosx = a \: cosecx + b \: secx \\ \\ \\ \hookrightarrow \tt a \: sinx - a \: cosecx + b \: cosx - b \: secx = 0 \\ \\ \\ \\ \hookrightarrow \tt \: a(sinx - \frac{1}{sinx} ) + b(cosx - \frac{1}{cosx} ) = 0 \\ \\ \\ \\ \hookrightarrow \tt \: \frac{ - a \: {cos}^{2} x}{ {sin}^{2}x } - \frac{b \: {sin}^{2}x }{cosx} = 0 \\ \\ \\ \\ \hookrightarrow \tt \: - a \: {cos}^{3} x - b {sin}^{3} x = 0 \\ \\ \\ \\ \hookrightarrow \tt \: - a \: cos {}^{3} x \: = b \: {sin}^{3} x \\ \\ \\ \\ \hookrightarrow \tt \: tan {}^{3} x = - \frac{a}{b} \\ \\ \\\\ \tt \hookrightarrow tanx = - ( { \frac{a}{b} })^{ \frac{1}{3} }$

Now,

$\tt \implies \: a \: sinx+ b \: cosx = cosx(a \: tanx + b) \\ \\ \\ = \tt \frac{1}{ \sqrt{1 + \frac{ {a}^{ \frac{2}{3} } }{ {b}^{ \frac{2}{3} } } } } ( \frac{ { - a}^{ \frac{4}{3} } + {b}^{ \frac{4}{3} } }{ {b}^{ \frac{1}{3} } } ) \\ \\ \\ = \tt \frac{ - {a}^{ \frac{4}{3} } + {b}^{ \frac{4}{3} } }{ \sqrt{ {a}^{ \frac{2}{3} } + {b}^{ \frac{2}{3} } } } \\ \\ \\ = \tt \: \frac{( {b}^{ \frac{2}{3} } + {a}^{ \frac{2}{3} })( { {b}^{ \frac{2}{3} } } - {a}^{ \frac{2}{3} } ) }{ \sqrt{ {b}^{ \frac{2}{3} } + {a}^{ \frac{2}{3} } } } \\ \\ \\ \implies \bigstar \boxed{ \tt \: {a}^{ \frac{2}{3} } - {b}^{ \frac{2}{3} } \sqrt{ {a}^{ \frac{2}{3} } + {b}^{ \frac{2}{3} } } }$

Hence proved !

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