if sin ( x - y) = 1/2 and cos (x + y) = 1/2, then what is the value of sinx cos x - 1 sin² x - cos² x sec x ?

- \(\frac{1}{√2}\)

We are given that :-

sin (x-y) =  \(\frac{1}{2}\)

{ we know, sin 30º =  \(\frac{1}{2}\) }

So,  ( x - y) = 30º   ---(1)

&  cos (x + y) =  \(\frac{1}{2}\)

{ we know, cos 60º =  \(\frac{1}{2}\) }

So, ( x + y ) = 60º     ----(2)

On adding equation 1 and 2.

2x = 90º

x = 45º

Now,

sinx cos x - 1 sin² x - cos² x sec x

= sin45º. cos45º - 1 sin²45º - cos²45º sec 45º

= \(\frac{1}{√2}\) × \(\frac{1}{√2}\) - \(\frac{1}{2}\) - \(\frac{√2}{2}\)

= \(\frac{1}{2}\)  - \(\frac{1}{2}\) - \(\frac{√2}{2}\)

= - \(\frac{1}{√2}\)