Question

if cosecA-sinA=a^3, secA-cosA=b^3, then prove that a^2b^2(a^2+b^2)=1

Answer 1

a^3=cosecA-sinA =>a^2=cosecA-sinA/a

b^3=secA-cosA =>b^2=secA-cosA/b

Now,

a^2=(1/sinA-sinA)/a

=(1-sin^2 A/sinA)/a

=(cos^2 A/sinA)/a

=cotA*cosA/a

b^2=secA-cosA/b

=(1/cosA-cosA)/b

=(1-cos^2 A/cosA)/b

=(sin^2 A/cosA)/b

=tanA*sinA/b

We have to find the value of


a^2xb^2(a^2+b^2)

So lets solve it

=>cotA*cosA/a . tanA*sinA/b{(cotA*cosA/a)+(tanA*sinA/b)}

=>cosA*sinA/ab . {cos^2 A/sinAa +sin^2 A/cosAb }

=>cosA*sinA/ab . {cos^2 A+sin^2 A/sinA.cosA.ab}

=>cosA*sinA/ab . {1/sinAcosAab} [sin^2 A + cos^2 A = 1]

=>1/(ab)2