Question

if x= cosec@[email protected] y=sec@-cos@ then prove that x⅔+y⅔=(xy)^-⅔​

Answer 1

Step-by-step explanation:

$x = cosec \theta - sin\theta \\ \\ y = sec\theta - cos\theta \\ \\ x = \frac{1}{sin\theta} - sin\theta \\ \\ x = \frac{1 - {sin}^{2} \theta}{sin\theta } \\ \\ x = \frac{ {cos}^{2}\theta }{sin\theta} \\ \\ x = \frac{ {cos}^{3} \theta}{sin\theta \: cos\theta} \\ \\ \\ \\ y = \frac{1}{cos\theta} - cos\theta \\ \\ y = \frac{1 - {cos}^{2} \theta}{cos\theta} \\ \\ y = \frac{ {sin}^{2} \theta}{cos\theta} \\ \\ y = \frac{ {sin}^{3}\theta }{sin\theta \: cos\theta} \\ \\ xy = sin\theta \: cos\theta \\ \\ x = \frac{ {cos}^{3}\theta }{xy} \\ \\ {x}^{2} y = {cos}^{3} \theta \: \: \: \: \: \: \: eq1\\ \\ y = \frac{ {sin}^{3} \theta}{xy} \\ \\ x {y}^{2} = {sin}^{3} \theta \: \: \: \: \: \: \: eq2\\ \\ {eq1}^{ \frac{2}{3} } + {eq2}^{\frac{2}{3} } \\ \\( {x}^{2} y) {}^{ \frac{2}{3} } + (x {y}^{2} ) {}^{ \frac{2}{3} } = {cos}^{2} \theta + {sin}^{2} \theta \\ \\ ( {x}^{2} y) {}^{ \frac{2}{3} } + (x {y}^{2} ) {}^{ \frac{2}{3} } =1 \\ \\ x {}^{ \frac{2}{3} } (xy) {}^{ \frac{2}{3} } + y {}^{ \frac{2}{3} } (xy) {}^{ \frac{2}{3} } = 1 \\ \\ (xy) {}^{ \frac{2}{3} } ( {x}^{ \frac{2}{3} } + {y}^{ \frac{2}{3} } ) =1 \\ \\ {x}^{ \frac{2}{3} } + {y}^{ \frac{2}{3} } = (xy) {}^{ - \frac{2}{3} }$

$formulae \\ \\ sin \theta \: cosec\theta = 1 \\ \\ cos\theta \: sec\theta = 1 \\ \\ {cos}^{2} \theta + {sin}^{2} \theta = 1$

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Answer 2

Step-by-step explanation:

$\begin{lgathered}x = cosec \theta - sin\theta \\ \\ y = sec\theta - cos\theta \\ \\ x = \frac{1}{sin\theta} - sin\theta \\ \\ x = \frac{1 - {sin}^{2} \theta}{sin\theta } \\ \\ x = \frac{ {cos}^{2}\theta }{sin\theta} \\ \\ x = \frac{ {cos}^{3} \theta}{sin\theta \: cos\theta} \\ \\ \\ \\ y = \frac{1}{cos\theta} - cos\theta \\ \\ y = \frac{1 - {cos}^{2} \theta}{cos\theta} \\ \\ y = \frac{ {sin}^{2} \theta}{cos\theta} \\ \\ y = \frac{ {sin}^{3}\theta }{sin\theta \: cos\theta} \\ \\ xy = sin\theta \: cos\theta \\ \\ x = \frac{ {cos}^{3}\theta }{xy} \\ \\ {x}^{2} y = {cos}^{3} \theta \: \: \: \: \: \: \: eq1\\ \\ y = \frac{ {sin}^{3} \theta}{xy} \\ \\ x {y}^{2} = {sin}^{3} \theta \: \: \: \: \: \: \: eq2\\ \\ {eq1}^{ \frac{2}{3} } + {eq2}^{\frac{2}{3} } \\ \\( {x}^{2} y) {}^{ \frac{2}{3} } + (x {y}^{2} ) {}^{ \frac{2}{3} } = {cos}^{2} \theta + {sin}^{2} \theta \\ \\ ( {x}^{2} y) {}^{ \frac{2}{3} } + (x {y}^{2} ) {}^{ \frac{2}{3} } =1 \\ \\ x {}^{ \frac{2}{3} } (xy) {}^{ \frac{2}{3} } + y {}^{ \frac{2}{3} } (xy) {}^{ \frac{2}{3} } = 1 \\ \\ (xy) {}^{ \frac{2}{3} } ( {x}^{ \frac{2}{3} } + {y}^{ \frac{2}{3} } ) =1 \\ \\ {x}^{ \frac{2}{3} } + {y}^{ \frac{2}{3} } = (xy) {}^{ - \frac{2}{3} } \\ \\\end{lgathered}$

$\begin{lgathered}formulae \\ \\ sin \theta \: cosec\theta = 1 \\ \\ cos\theta \: sec\theta = 1 \\ \\ {cos}^{2} \theta + {sin}^{2} \theta = 1 \\ \\\end{lgathered}$