1.

Find value of sinx and cosy​

Answer»

Answer:

=>>>2^(sinx+cosy)=1

2^(sinx+cosy)=2^0

sinx+cosy=0

Sinx=-cosy

=>>>16^(sin^2x+cos^y)=4

4^2(sin^2x+cos^2y)=4

2(sin^2x+cos^2y)=1

2(2*cos^2y)=1

Cos^2y=1/4

Cosy=1/2

Sinx=-1/2



Discussion

No Comment Found

Related InterviewSolutions