Question

Dare to solve this question
$\sf if \: cosec \theta - sin \theta = {a}^{3} \: and \: sec \theta - cos \theta = {b}^{3}$
prove that a²b²(a²+b²)=1​

Answer 1

Step-by-step explanation:

see this image you will get it

Answer 2

$\pink\bigstar$SOLUTION:-

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Given that,

$\sf \star cosec \theta - sin \theta = {a}^{3}$

Now using below trignometric identity:-

$\longrightarrow \bf\dfrac{1}{cosec \theta} = sin \theta$

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We have ,

$\implies \sf\dfrac{1}{sin \theta} - sin \theta = {a}^{3}$

$\implies \sf \dfrac{ 1 - sin^{2} \theta }{sin \theta} = {a}^{3}$

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Using below trignometric identity:-

$\bf \longrightarrow1 - sin ^{2} \theta = cos^{2} \theta$

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We have ,

$\implies \sf\dfrac{cos^{2} \theta }{sin \theta} = {a}^{3}$

$\implies \sf a = \sqrt{\dfrac{(cos \theta)^{2} }{sin \theta} }$

$\implies \bf a = \sqrt{ \dfrac{(cos \theta)^{ \frac{2}{3} } }{(sin \theta) ^{ \frac{1}{3} } } } .....(1)$

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Again , Given that,

$\sf \star sec \theta - cos \theta = {b}^{3}$

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Using below trignometric identity

$\bf \longrightarrow sec \theta = \dfrac{1}{cos \theta}$

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We have,

$\implies \sf \dfrac{1}{cos \theta} - cos \theta = {b}^{3}$

$\implies \sf \dfrac{1 - cos ^{2} \theta }{cos \theta} = {b}^{3}$

Using below trignometric identity

$\bf \longrightarrow 1 - {cos}^{2} \theta = {sin}^{2} \theta$

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We have

$\implies \sf \dfrac{sin^{2} \theta }{cos \theta} = {b}^{3}$

$\implies \sf b = \sqrt{ \dfrac{(sin \theta)^{2} }{cos \theta} }$

$\implies \bf b = \dfrac{(sin \theta)^{ \frac{2}{3} } }{(cos \theta)^{ \frac{1}{3} } } ......(2)$

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Now , LHS=a²b²(a²+b²)

Using (1) and (2) we have

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$\bf = \dfrac{(cos \theta)^{ \frac{4}{3} } }{(sin \theta)^{ \frac{2}{3} } } \times \dfrac{(cos \theta)^{ \frac{4}{3} } }{(sin \theta)^{ \frac{2}{3} } } \bigg \{\dfrac{(cos \theta)^{ \frac{4}{3} } }{(sin \theta)^{ \frac{2}{3} } } + \dfrac{(cos \theta)^{ \frac{4}{3} } }{(sin \theta)^{ \frac{2}{3} } } \bigg \}$

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$\bf = (cos \theta) ^{\frac{2}{3} } \times (sin \theta) ^{ \frac{2}{3} } \bigg \{ \dfrac{(cos \theta)^{2} + (sin \theta)^{2} }{(sin \theta)^{ \frac{2}{3} } \times (cos \theta)^{ \frac{2}{3} } }\bigg\}$

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$\bf = \dfrac{ \cancel{(cos \theta) ^{ \frac{2}{3} } .(sin \theta)^{ \frac{2}{3}} } .(cos^{2} \theta + sin^{2} \theta) }{ \cancel{(cos \theta) ^{ \frac{2}{3} } .(sin \theta)^{ \frac{2}{3} }} }$

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$= \bf(cos^{2} \theta + sin ^{2} \theta)$

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$= 1$

Hence,RHS=LHS Proved