The expression cos^2(A - B) + cos^2 B - 2cos( | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) Let the expression be $E = \cos^2(A - B) + \cos^2 B - 2\cos(A - B)\cos A\cos B$.Using the identity $2\cos A\cos B = \cos(A - B) + \cos(A + B)$,we

Vedclass · Accepted Answer

Dependent on $A$

Vedclass · Answer

(C) Let the expression be $E = \cos^2(A - B) + \cos^2 B - 2\cos(A - B)\cos A\cos B$.Using the identity $2\cos A\cos B = \cos(A - B) + \cos(A + B)$,we get:$E = \cos^2(A - B) + \cos^2 B - \cos(A - B)[\cos(A - B) + \cos(A + B)]$$E = \cos^2(A - B) + \cos^2 B - \cos^2(A - B) - \cos(A - B)\cos(A + B)$$E = \cos^2 B - \cos(A - B)\cos(A + B)$Using the identity $\cos(A - B)\cos(A + B) = \cos^2 A - \sin^2 B$,we get:$E = \cos^2 B - (\cos^2 A - \sin^2 B)$$E = \cos^2 B + \sin^2 B - \cos^2 A$Since $\cos^2 B + \sin^2 B = 1$,we have:$E = 1 - \cos^2 A = \sin^2 A$.Since the result $\sin^2 A$ depends only on $A$,the expression is dependent on $A$.

The expression $\cos^2(A - B) + \cos^2 B - 2\cos(A - B)\cos A\cos B$ is

Similar Questions

If $\cos A = \frac{3}{4}$,then $32\sin \frac{A}{2}\cos \frac{5A}{2} = $

If $\sec \theta + \tan \theta = m$ $(m > 1)$,then the value of $\sin \theta$ is $(0^{\circ} < \theta < 90^{\circ})$.

If $\cot 20^{\circ} = p$,then $\frac{\tan 160^{\circ} - \tan 110^{\circ}}{1 + \tan 160^{\circ} \tan 110^{\circ}} =$

If $a \tan \theta = b$,then $a \cos 2\theta + b \sin 2\theta = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module