If cos A = frac{4}{5},then the value of tan A | vedclass

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

Question

(A) Given,$\cos A = \frac{4}{5}$.We know that $\sin^2 A + \cos^2 A = 1$,so $\sin A = \sqrt{1 - \cos^2 A}$.Substituting the value of $\cos A$:$\sin A =

Vedclass · Accepted Answer

$\frac{3}{4}$

Vedclass · Answer

(A) Given,$\cos A = \frac{4}{5}$.We know that $\sin^2 A + \cos^2 A = 1$,so $\sin A = \sqrt{1 - \cos^2 A}$.Substituting the value of $\cos A$:$\sin A = \sqrt{1 - (\frac{4}{5})^2} = \sqrt{1 - \frac{16}{25}} = \sqrt{\frac{25 - 16}{25}} = \sqrt{\frac{9}{25}} = \frac{3}{5}$.Now,$	an A = \frac{\sin A}{\cos A}$.Substituting the values of $\sin A$ and $\cos A$:$	an A = \frac{3/5}{4/5} = \frac{3}{4}$.Thus,the value of $	an A$ is $\frac{3}{4}$.

If $\cos A = \frac{4}{5}$,then the value of $\tan A$ is

Similar Questions

In $\Delta ABC$,$m \angle C = 90^{\circ}$ and $\tan A = \frac{1}{\sqrt{3}}$,then $\sin A = \ldots$

$\sin \theta \cdot \cos (90^\circ - \theta) = \ldots \ldots \ldots$

Write 'True' or 'False' and justify your answer.
The value of the expression $(\cos^{2} 23^{\circ} - \sin^{2} 67^{\circ})$ is positive.

If $a \sin \theta + b \cos \theta = c$,then prove that $a \cos \theta - b \sin \theta = \pm \sqrt{a^2 + b^2 - c^2}$,given $a^2 + b^2 \geq c^2$.

$\sin (90^\circ - \theta) = \ldots \ldots \ldots$

Vedclass Test Series

Exam Paper Generator

Online Exam Module