Question

The value of
sin(n+1) A – sin(n-1) A
cos(n+1) A + cos(n-1) A
is​

Answer 1

Answer:

As,

cos(A-B) = cosAcosB+sinAsinB....(1)

Now let,

A= (n+1)A

B=(n-1)A

then by eq (1) we can say that

cos[(n+1)A - (n-1)A]= sin(n+1)A.sin(n-1)A+cos(n+1)A.cos(n-1)A

in lhs,

=cos[An+A-(An-A)]

=cos 2A

hence proved

hope it helps uh...

Step-by-step explanation:

MissInvisible...✌️