[" (iv) "cos(A-B)=cos A cos B+sin A sin D],[" (v) "tan(A-B)=(tan A-tan B)/(1+tan A tan B)]

If A=B=60o, verify that cos(A-B)=cos A cos B+sin A sin B(iii)sin(A-B)=sin A cos B-cos A sin B(iii)tan(A-B)=(tan A-tan B)/(1+tan A tan B)

By using above basic addition/ subtraction formulae, prove that (i) tan (A+B)=(tan A+tan B)/(1-tan A tan B) , (ii) tan (A-B)=(tan A-tanB)/(1+tan A tan B) (iii) sin2theta=2sintheta costheta , (iv) cos2theta=cos^(2)theta=1-2sin^(2)theta=2cos^(2)theta-1 (v) tan 2theta=(2 tan theta)/(1-tan^(2) theta)

By using above basic addition/ subtraction formulae, prove that (i) tan (A+B)=(tan A+tan B)/(1-tan A tan B) , (ii) tan (A-B)=(tan A-tanB)/(1+tan A tan B) (iii) sin2theta=2sintheta costheta , (iv) cos2theta=cos^(2)theta=1-2sin^(2)theta=2cos^(2)theta-1 (v) tan 2theta=(2 tan theta)/(1-tan^(2) theta)

By using above basic addition/ subtraction formulae, prove that (i) tan (A+B)=(tan A+tan B)/(1-tan A tan B) , (ii) tan (A-B)=(tan A-tanB)/(1+tan A tan B) (iii) sin2theta=2sintheta costheta , (iv) cos2theta=cos^(2)theta=1-2sin^(2)theta=2cos^(2)theta-1 (v) tan 2theta=(2 tan theta)/(1-tan^(2) theta)

If (x)/(y)=(cos A)/(cos B) where A!=B ,then (a) tan(A+B)/(2)=(x tan A+y tan B)/(x+y) (b) tan(A-B)/(2)=(x tan A-y tan B)/(x+y) (c) (sin(A+B))/(sin(A-B))=(y sin A+x sin B)/(y sin A-x sin B) (d) (x cos B-y cos A)/(xcosB+ ycosA)=0

If A + B + C = 180^(@) prove that: (i) "sin" (B+C-A) + "sin"(C+ A-B) + sin (A + B-C) = 4 sin A sin B sin C (ii) cos(- A+B+C) + cos (A - B + C) + cos (A + B - C) = 1 + 4 cos A cos B cos C (iii) "tan" (B+C - A) + tan (C+A-B) + tan (A + B-C) = tan (B+C - A) tan (C+ A - B) tan (A+B-C)

If cos(A+B) sin (C-D) = cos (A-B) sin (C+D) , then show that tan A tan B tan C + tan D =0

If cos(A+B) sin (C-D) = cos (A-B) sin (C+D) , then show that tan A tan B tan C + tan D =0

If sin A = (3)/(5) and cos B = (12)/(13) . Then the value of (tan A - tan B)/(1 + tan A tan B) is equal to

If sin A cos B = (1)/(4) and 3 tan A = tan B then triangle is