If A = $60^{\circ} a n d B = 60^{\circ},$ verify that:

(i) sin (A - B) = sin A cos B - cos A sin B

(ii) cos (A - B) = cos A cos B + sin A sin B

(iii) tan (A - B) = $\frac{t a n A - t a n B}{1 + t a m A t a n B}$

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Solution

Given A=60° and B=60°

taking LHS
SIN(60-60) = sin 0°=0

solving RHS
sin A cos B - cos A sin B
=sin 60°cos60° - cos 60°sin60°

taking sin60° common

sin60°(cos 60°-cos 60°) = 0 = LHS
hence verified

(ii)taking LHS
cos (A - B) = cos(60-60) = cos 0 = 1
cos A cos B + sin A sin B

= Cos 60°cos 60° + sin 60°sin 60°

RHS = LHS

hence verified

(iii)TAKING LHS

tan (A - B) =tan (60-60)

= tan 0° = 0

SOLVING RHS
$fraction numerator tan space 60 space minus tan space 60 over denominator 1 plus tan 60 space tan space 60 end fraction fraction numerator 0 over denominator 1 plus tan space 60 space tan 60 end fraction equals 0 equals L H S h e n c e space v e r i f i e d$

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If A = $60^{\circ} a n d B = 60^{\circ},$ verify that:

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(ii) cos (A - B) = cos A cos B + sin A sin B

(iii) tan (A - B) = $\frac{t a n A - t a n B}{1 + t a m A t a n B}$