Question

If cos (x - y) = a cos (x + y), then cot x cot y is equal to​

Answer 1

Given,

cos(x-y) = a cos(x+y) ;

ie,

cos(x)cos(y)+sin(x)sin(y) =a[cos(x)cos(y) - sin(x)sin(y)] ;

ie,

cos(x)cos(y)+sin(x)sin(y)

=acos(x)cos(y) - asin(x)sin(y);

ie,

sin(x)sin(y)+asin(x)sin(y)

=acos(x)cos(y)-cos(x)cos(y);

Now we have,

(1+a)sin(x)sin(y) = (a-1)cos(x)cos(y);

ie,

cos(x)cos(y)÷sin(x)sin(y) =(1+a)÷(a-1);

(since, we know that cosx ÷ sinx = cotx),

We now have,

cotx coty = (1+a)÷(a-1)

or

cotx coty = (1+a)/(a-1)