Question #6ad0a

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Question #6ad0a

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Answer 1 · 2017-01-19T08:20:45

Given $csc x - sin x = a$ $cscx-sinx=a$

$\Rightarrow \frac{1}{sin x} - sin x = a$ $=>1/sinx-sinx=a$

$\Rightarrow \frac{1 - {sin}^{2} x}{sin x} = a$ $=>(1-sin^2x)/sinx=a$

$\Rightarrow \frac{{cos}^{2} x}{sin x} = a$ $=>cos^2x/sinx=a$

Again $sec x - cos x = b$ $secx-cosx=b$

$\Rightarrow \frac{1}{cos x} - cos x = b$ $=>1/cosx-cosx=b$

$\Rightarrow \frac{1 - {cos}^{2} x}{cos x} = b$ $=>(1-cos^2x)/cosx=b$

$\Rightarrow \frac{{sin}^{2} x}{cos x} = b$ $=>sin^2x/cosx=b$

Now $a^{2} b^{2} (a^{2} + b^{2} + 3)$ $a^2b^2(a^2+b^2+3)$

$= \frac{{cos}^{4} x}{{sin}^{2} x} \times \frac{{sin}^{4} x}{{cos}^{2} x} (\frac{{cos}^{4} x}{{sin}^{2} x} + \frac{{sin}^{4} x}{{cos}^{2} x} + 3)$ $=cos^4x/sin^2x xxsin^4x/cos^2x(cos^4x/sin^2x+sin^4x/cos^2x+3)$

$= {sin}^{2} x \times {cos}^{2} x (\frac{{cos}^{6} x + {sin}^{6} x}{{sin}^{2} x {cos}^{2} x} + 3)$ $=sin^2x xxcos^2x((cos^6x+sin^6x)/(sin^2xcos^2x)+3)$

$= {cos}^{6} x + {sin}^{6} x + 3 {sin}^{2} x {cos}^{2} x$ $=cos^6x+sin^6x+3sin^2xcos^2x$

$= {({sin}^{2} x)}^{3} + {({cos}^{2} x)}^{3} + 3 {sin}^{2} x {cos}^{2} x ({sin}^{2} x + {cos}^{2} x)$ $=(sin^2x)^3+(cos^2x)^3+3sin^2xcos^2x(sin^2x+cos^2x)$

$= {({sin}^{2} x + {cos}^{2} x)}^{3} = 1^{3} = 1$ $=(sin^2x+cos^2x)^3=1^3=1$

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Question #6ad0a

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