Question

Please anyone solve this question.​

Answer 1

Answer:

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Step-by-step explanation:

$To \: Prove = \\ \frac{1}{cosec \: θ + cot \: θ} - \frac{1}{sin \:θ } = \frac{1}{sin \:θ} - \frac{1}{cosec \:θ - cot \:θ \ } \\ \\ ie \: \: \frac{1}{cosec \:θ + cot θ} + \frac{1}{cosec \: θ - cot \:θ} = \frac{1}{sin \:θ } + \frac{1}{sin \: θ} \\ \\ let \: \: LHS = \frac{1}{cosec \:θ + cot \: θ } + \frac{1}{cosec \:θ - cot \:θ } \\ \\ RHS = \frac{1}{sin \: θ} + \frac{1}{sin \: θ}$

$considering \: LHS \: first \\ = \frac{1}{cosec \: θ + cot \:θ } + \frac{1}{cosec \: θ - cot \:θ } \\ \\ = \frac{cosec \:θ + cot \: θ + cosec \: θ - cot \:θ }{(cosec \:θ + cot \:θ)(cosec \: θ - cot \: θ)} \\ \\ = \frac{2.cosec \:θ }{cosec {}^{2} θ - cot {}^{2} θ} \\ \\ = \frac{2.cosec \:θ }{1} \\ since \: cosec {}^{2} θ - cot {}^{2} θ = 1 \\ \\ thus \: LHS = 2.cosec \: θ$

$now \: considering \: RHS \\ = \frac{1}{sin \: θ} + \frac{1}{sin \: θ} \\ \\ = \frac{sin \:θ + sin \: θ }{sin \: θ.sin \:θ} \\ \\ = \frac{2.sin \: θ}{sin {}^{2} θ} \\ \\ = \frac{2}{sin \:θ } \\ \\ = 2 \times \frac{1}{sin \:θ } \\ \\ = 2.cosec \: θ \\ since \: \: \frac{1}{sin \: θ} = cosec \: θ \\ \: \\ hence \: RHS = 2.cosec \: θ \\ \\ thus \: LHS=RHS \\ hence \: proved$