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

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

Question

(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 $\cos (A + B) = \dots$

Similar Questions

$\frac{1}{\tan ^{2} \theta}+1 = \dots$

If $\sec 4A = \operatorname{cosec}(A - 20^\circ)$,where $4A$ is an acute angle,then the value of $A$ is $\ldots \ldots \ldots \ldots .$ (in $^\circ$)

The value of the expression $\left[\operatorname{cosec}(75^{\circ}+\theta)-\sec(15^{\circ}-\theta)-\tan(55^{\circ}+\theta)+\cot(35^{\circ}-\theta)\right]$ is

Given that $\sin \alpha = \frac{1}{2}$ and $\cos \beta = \frac{1}{2}$,then the value of $(\alpha + \beta)$ is (in $^{\circ}$)

$\sin 60^{\circ} \sin 45^{\circ} + \cos 60^{\circ} \cos 45^{\circ} = \ldots \ldots \ldots \ldots$

Vedclass Test Series

Exam Paper Generator

Online Exam Module