Question

prove that cos(A+B)COS(A-B)=cos²A-sin²B​

Answer 1

Answer:

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Answer 2

Given :

$\cos(A+B)\cos(A-B)=\cos^2A-\sin^2B$

To Find :

$\text{Prove RHS = LHS}$

Identities Used :

$\cos(x+y)=\cos x \cos y - \sin x \sin y$

$\cos(x-y)=\cos x \cos y + \sin x \sin y$

$\cos ^2 = 1- \sin ^2x$

$\sin ^2x = 1 - \cos ^2 x$

Solution :

$\textbf {LHS :}$

$\cos ^2A - \sin^2 B$

$\textbf{RHS :}$

$\cos (A+B) \cos (A-B)$

$\implies (\cos A \cos B - \sin A \sin B )(\cos A \cos B + \sin A \sin B)$

$(\because \cos(A+B)=\cos A \cos B- \sin A \sin B )$

$(\because \cos(A-B)=\cos A \cos B+ \sin A\sin B )$

$\implies \cos^2 A \cos ^2 B - \sin ^2 A \sin^2 B$

$( \because a^2 - b^2 = (a+b)(a-b))$

$\implies \cos^2 A(1- \sin^2 B) - (1-\cos^2 A)\sin ^2 B$

$(\because \cos ^2 B = 1- \sin ^2B \: \: \& \:\: \sin ^2A = 1 - \cos ^2 A)$

$\implies (\cos ^2 A - \cos ^2 A \sin ^2 B )-(\sin ^2 B - \sin ^2 B \cos ^2 A)$

$\implies \cos ^2 A - \cos ^2 A \sin ^2 B - \sin ^2 B + \sin ^2 B \cos ^2 A$

$\implies \cos^2 A - \sin ^2 B = \bf LHS$

Hence Proved.