Let y= sqrt [1-sin 2x / 1+sin 2x] , prove that dy/dx + sec^2 (pi/4-x)=0

Question

This question is from isc class 12 book

 

Answer 1

Note cos(π/4 - x) = cos(π/4 ) * cos(x) + sin(π/4 ) * sin(x)

 

                            = (cos(x) + sin(x))/√(2)

 

cos²(π/4 - x) = (cos(x) + sin(x))²/2

                     = (cos²(x) + sin²(X) + 2*cos(x)*sin(x))/2

                     = (1 + sin(2x))/2

 

From this sec²(π/4 - x) = 2/(1 + sin(2x))

 

Also, 1 - sin(2x) = sin²(x) + cos²(x) - 2*sin(x)*cos(x)

                          = (cos(x) - sin(x))²

 

Similarly, 1 + sin(2x) = (cos(x) + sin(x))²

 

Also (1 - sin(2x))/(1 + sin(2x)) = ((cos(x) - sin(x))/(cos(x) + sin(x)))²

                                                = ((cos(x) - sin(x))*(cos(x) + sin(x))/(cos(x) + sin(x))²)²

                                                = ((cos²(x) - sin²(x))/(1 + sin(2x)))²

                                                = (cos(2x)/(1 + sin(2x)))²

 

Now y =√((1 - sin(2x))/(1 + sin(2x)))

 

          = √((cos(2x)/(1 + sin(2x)))²)

 

          = cos(2x)/(1 + sin(2x))

 

Take the derivative:

 

   y' = -2sin(2x)/(1 + sin(2x)) + cos(2x) (-1)(2cos(2x))/(1 + sin(2x))²

 

       = -2/(1 + sin(2x))²*(sin(2x)*(1 + sin(2x)) + cos²(2x))

 

       = -2/(1 + sin(2x))²*(sin(2x) + sin²(2x) + cos²(2x))

 

       = -2/(1 + sin(2x))²*(1 + sin(2x))

 

       = -2/(1 + sin(2x))

 

      = -sec²(π/4 - x), from the identity above

Answer 2

Let u = 1-sin2x & v = 1+sin2x

and let y = u^½v^-½ and

thus u'=-2cos2x, v'=2cos2x

Then by product rule,

y' = (d/dx)u^½•v^-½ + u^½•(d/dx)v^-½

....

=½(1-sin2x)^-½(-2cos2x)(1+sin2x)^-½ 

   -½(1-sin2x)^½(1+sin2x)^-(3/2)(2cos2x)

=(-cos2x)(1+sin2x)^-(3/2)(1-sin2x)^-½ ...factored, & times [(1+sin2x) + (1-sin2x)]

which I simplify to -2cos2x over

   (1+sin2x)(3/2)(1-sin2x)½

and that's as far as I wish to go! Check it very carefully because it was a nightmare to key in...and of course this is not the completed job.