Question

1/sec2x - cos2x+1/cosec2-sin2x × sin2xcos2x = 1-sin2xcos2/2+sin2xcos2x

Answer 1

Step-by-step explanation:

We have to prove that 1−sin2xcos2x=cos2x1+sin2x

To do this we transform left side:

   1−sin2xcos2x=(sin2x+cos2x)−2sinxcosxcos2x

   =sin2x−2sinxcosx+cos2xcos2x−sin2x

   =(sinx−cosx)2(cosx−sinx)(cosx+sinx)

   =(sinx−cosx)2−(sinx−cosx)(sinx+cosx)

   =cosx−sinxcosx+sinx

Now we expand the expresion by multiplying both numerator and denominator by (cosx+sinx)

So we get:

   (cosx−sinx)(cosx+sinx)(cosx+sinx)2

   =cos2x−sin2xcos2x+2cosxsinx+sin2x

   =cos2x1+sin2x

    i hope it is helpfull

    plz mark as brainlist answer.....