Question

if cos(pi/4 -x)cos2x + sinx sin2x secx = cos sin2x secx + cos(pi/4 + x)cos2x then possible value of cos^2x + sec^2x is​

Answer 1

Step-by-step explanation:

cos(pi/4 -x)cos2x + sinx sin2x secx = cos sin2x secx + cos(pi/4 + x)cos2x

cos(pi/4 -x) = cospi/4cosx + sinpi/4sinx

= 1/root2 cosx + 1/root2 sinx

= 1/root2 (cosx + sinx)

sinx sin2x secx = sinx 2sinxcosx (1/ cosx)

=2(sinx)^2

cosxsin2x secx = cosx sin2x (1/cosx) = sin2x

cos(pi/4 + x) = cospi/4cosx - sinpi/4sinx

= 1/root2 cosx - 1/root2 sinx

= 1/root2 (cosx - sinx)

cos(pi/4 -x)cos2x + sinx sin2x secx = cos sin2x secx + cos(pi/4 + x)cos2x

1/root2 (cosx + sinx) cos2x+ 2(sinx)^2 = sin2x + 1/root2 (cosx - sinx)cos2x

1/root2 (cosx + sinx -cosx + sinx)cos2x=2sinxcosx - 2(sinx)^2

1/root2(2sinx)cos2x= 2sinx(cosx-sinx)

cos2x/(cosx-sinx) = root2

(cos^2x-sin^2x)/(cosx-sinx) = root2

(cosx-sinx)(cosx+sinx)/cosx-sinx) = root2

cosx+sinx= root2

cosx+sinx=root2

(cosx+sinx)^2=(root2)^2

cos^2x+sin^2x+2sinxcosx=2

1+sin2x=2

sin2x=2-1

sin2x=1

sin90=1

2x=90

x=45

cosx=cos45 =1/root2

secx=sec45=root2

cos^2x+sec^2x = (1/root2)^2+(root2)^2

=1/2+2

=5/2