Question

If sin A=√3/2=cos B then find tan(A-B)​

Answer 1

sin A = √3/2

cos B = √3/2

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

cos A = - √(1-sin^2 A)

= - √(1- (√3/2)^2 )

= - √(1- 3/4 )

= - √1/4 = - 1/2

sin B = √(1- cos^2 A)

= √ (1 - (-1/2)^2)

= √( 1 - 1/4)

= √3/2

• tan A = sinA/cosA = (√3/2) / (-1/2) = - √3
• tan B = sinB/cosB = (√3/2) / (√3/2) = 1

tan ( A-B) = (-√3 - 1) / ( 1+ (-√ 3)(1) )

= (-√3 - 1)/(1-√3)

= (√3+1) / (√3-1)

note: also doing the same answer in attachment...