T.R of Compound Angles#!#Sin(A+B),Cos(A+B)...

Trigonometric ratios OF compound angle Sin(A+B) || Sin(A-B) || Cos(A+B) || Cos(A-B) and problem on these

If A, B, C, D are four angles then sin (AB) .cos (A + B) + sin (BC) .cos (B + C) + sin (CD) .cos (C + D) + sin ( DA) .cos (D + A) =

For all values of angle A and B(i)sin(A-B)=sin A cos B-cos A sin B(ii)sin(A+B)=sin A cos B+cos A sin B

If A and B are complementary angles,then sin A=sin B(b)cos A=cos B(c)tan A=tan B(d)sec A=cos ecB

Prove that formula cos(A-B)=cos A*cos B+sin A*sin Bwith the help of the formula sin(A+B)=sin A*cos B+cos A*sin B

Prove that (1 + sin A) / (cos A) + (cos B) / (1-sin B) = (2sin A-2sin B) / (sin (AB) + cos A-cos B)

Standard Formula sin(A+B) || sin(A–B) || cos(A+B) || cos(A–B)

Standard Formula sin(A+B) || sin(A–B) || cos(A+B) || cos(A–B)

Standard Formula sin(A+B) || sin(A–B) || cos(A+B) || cos(A–B)