Question #6ad0a

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Answer 1 · 2017-01-19T08:20:45

Given $cscx-sinx=a$

$=>1/sinx-sinx=a$

$=>(1-sin^2x)/sinx=a$

$=>cos^2x/sinx=a$

Again $secx-cosx=b$

$=>1/cosx-cosx=b$

$=>(1-cos^2x)/cosx=b$

$=>sin^2x/cosx=b$

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

$=cos^4x/sin^2x xxsin^4x/cos^2x(cos^4x/sin^2x+sin^4x/cos^2x+3)$

$=sin^2x xxcos^2x((cos^6x+sin^6x)/(sin^2xcos^2x)+3)$

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

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

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

Proved