If `A=B=60^@` , verify that (i) `cos(A-B)=cosAcosB+sinAsinB` (ii) `sin(A-B)=sinAcosB-cosAsinB` (iii) `tan(A-B)=(tanA-tanB)/(1+tanAtanB)`

Explanation

Given ` A=B=60^@`

`Cos(A-B)=Cos(0)=1.`

Now, `CosACosB+SinASinB=Cos(60^@)Cos(60^@)+Sin(60^@)Sin(60^@)`

=>`3/4 +1/4=4/4=1`

henc proved.

Second Part,

`tan(A-B)=( tanA-tanB)/(1+tanAtanB)`

`tan(A-B)=tan(60-60)=tan0=0`
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