For acute angles A and B, if tan A = 1 and si | vedclass

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(A) Given that $A$ and $B$ are acute angles.Since $	an A = 1,$ we know that $	an 45^{\circ} = 1,$ therefore $A = 45^{\circ}.$Since $\sin B = \frac{1

Vedclass · Accepted Answer

$0$

Vedclass · Answer

(A) Given that $A$ and $B$ are acute angles.Since $	an A = 1,$ we know that $	an 45^{\circ} = 1,$ therefore $A = 45^{\circ}.$Since $\sin B = \frac{1}{\sqrt{2}},$ we know that $\sin 45^{\circ} = \frac{1}{\sqrt{2}},$ therefore $B = 45^{\circ}.$Now,we need to find the value of $\cos (A + B).$Substituting the values of $A$ and $B,$ we get $\cos (45^{\circ} + 45^{\circ}) = \cos 90^{\circ}.$Since $\cos 90^{\circ} = 0,$ the final answer is $0$.

For acute angles $A$ and $B,$ if $\tan A = 1$ and $\sin B = \frac{1}{\sqrt{2}},$ then $\cos (A + B) = \dots$

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