If A = B = 60°, verify that
(i)cos (A − B) = cos A cos B + sin A sin B
(ii)sin (A − B) = sin A cos B − cos A sin B
(iii)
Answers
SOLUTION :
(i) Given : A = B = 60°
verify : Cos (A – B) = Cos A Cos B + Sin A Sin B
L.H.S = Cos (A – B)
On Substituting A = B = 60° in LHS
= cos (60° – 60°)
= cos 0°
= 1
[cos 0°= 1]
L.H.S = 1
R.H.S = Cos A Cos B + Sin A Sin B
On Substituting A = B = 60° in RHS
= cos 60° cos 60° + sin 60° sin 60°
= cos² 60° + sin² 60°
= (½)² + (√3/2)²
[cos 60° = ½ , sin 60° = √3/2]
= (¼ ) + 3/4
= (1+3)/4
= 4/4 = 1
= 1
RHS = 1
LHS = RHS = 1
Hence, Cos (A – B) = Cos A Cos B + Sin A Sin B
(ii) Given : A = B = 60°
verify : sin (A − B) = sin A cos B − cos A sin B
L.H.S = sin (A − B)
On Substituting A = B = 60° in LHS
= sin (60° – 60°)
= sin 0°
= 0
[sin 0°= 0]
L.H.S = 0
R.H.S = sin A cos B − cos A sin B
On Substituting A = B = 60° in RHS
= sin 60° cos 60° - cos 60° sin 60°
= √3/2 × ½ - ½ × √3/2
[cos 60° = ½ , sin 60° = √3/2]
= √3/4 - √3/4
= 0
RHS = 0
LHS = RHS = 0
Hence, sin (A − B) = sin A cos B − cos A sin B
(iii) Given : A = B = 60°
Verify : tan(A−B) = tan A −tanB / 1+ tanA tanB
LHS = tan(A−B)
On Substituting A = B = 60° in LHS
= tan(60° − 60°)
= tan 0°
= 0
[ tan 0° = 0 ]
LHS = 0
RHS = tan A −tanB / 1+ tanA tanB
= tan 60° − tan 60° / 1+ tan 60° tan 60°
= 0 /1 + tan² 60°
= 0/ 1 + (√3)²
= 0/ 1 + 3
= 0 /4 = 0
= 0
RHS = 0
LHS = RHS = 0
Hence, tan(A−B) = tan A −tanB / 1+ tanA tanB
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