if cosec theta -sin theta = a cube and sec theta -cos theta = b cube then prove that a square.b sq(a sq + b sq) = 1

so here it is ..

cosecÓ¨ - sinÓ¨ = a³...

or (1 sin²Ó¨) /sinÓ¨ = cos²Ó¨ / sinÓ¨ = a³...

or a = (cos²Ó¨ / sinÓ¨)^2/3 (i)

and

secÓ¨ cosÓ¨ = b³

or (1 cos²Ó¨) / cosÓ¨ = sin²Ó¨ / cosÓ¨ = b³

or b = (sin²Ó¨/cosÓ¨)^2/3.. (ii)

Now lets put the last nail in this question's coffin Jby putting the values of a and b in the equation..

..a²b²(a²+ b²) ..

=(cos²Ó¨ / sinÓ¨)^2/3(sin²Ó¨/ cosÓ¨)^2/3 {(cos²Ó¨ / sinÓ¨)^2/3+(sin²Ó¨/ cosÓ¨)^2/3}

={ (cos²Ó¨ / sinÓ¨)(sin²Ó¨/ cosÓ¨)(cos²Ó¨ / sinÓ¨) }^2/3+{ (cos²Ó¨ / sinÓ¨)(sin²Ó¨/ cosÓ¨) (sin²Ó¨ / cosÓ¨) }^2/3

={cosÓ¨sinÓ¨(cos²Ó¨ / sinÓ¨) }^2/3+{cosÓ¨ sinÓ¨(sin²Ó¨/ cosÓ¨) }^2/3

=(cos³Ó¨)^2/3+(sin³Ó¨)^2/3

=cos²Ó¨+ sin²Ó¨ = 1

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hey in the 3rd and 7th line there is a minus sign b/w 1 sin(sq) theta nd b/w 1 cos(sq)theta

nd 1 more thing eq(i) nd eq (ii) are fr a(sq) and b(sq)... nt fr 'a' nd 'b'.... nd then we put the values of a(sq) nd b(sq) in the eqna²b²(a²+ b²)... srry printing mistakes :P