Question

cos(A+B) cos ( A-B)=​

Answer 1

cos(a+b)*cos(a-b) = cos^2a - sin^2b tan(x/2) = (1-cosx)/(sinx)

Answer 2

Step-by-step explanation:

We have,,

$\sf{\cos(A+B).\cos(A-B)}$

$\sf{=[\cos(A).\cos(B)-\sin(A).\sin(B)].[\cos(A).\cos(B)+\sin(A).\sin(B)]}$

$\sf{=\cos^{2}(A).\cos^{2}(B)-\sin^{2}(A).\sin^{2}(B)}$

$\sf{=[1-\sin^{2}(A)].[1-sin^{2}(B)]-\sin^{2}(A).\sin^{2}(B)}$

$\sf{=[1-\sin^{2}(B)-\sin^{2}(A)+\sin^{2}(A).\sin^{2}(B)]-\sin^{2}(A).\sin^{2}(B)}$

$\sf{=1-\sin^{2}(B)-\sin^{2}(A)+\sin^{2}(A).\sin^{2}(B)-\sin^{2}(A).\sin^{2}(B)}$

$\sf{={\boxed{\sf{\cos^{2}(B)-\sin^{2}(A)}}}}$

Hope it helps ❣️❣️❣️