Prove the following

Question:

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

(a) $\frac{3}{5}$

(b) $\frac{3}{4}$

(C) $\frac{4}{3}$

(d) $\frac{5}{3}$

Solution:

(b) Given, $\cos A=\frac{4}{5}$

$\therefore \quad \sin A=\sqrt{1-\cos ^{2} A}$  $\left[\begin{array}{l}\because \sin ^{2} A+\cos ^{2} A=1 \\ \therefore \sin A=\sqrt{1-\cos ^{2} A}\end{array}\right]$

$=\sqrt{1-\left(\frac{4}{5}\right)^{2}}=\sqrt{1-\frac{16}{25}}=\sqrt{\frac{9}{25}}=\frac{3}{5}$

Now, $\tan A=\frac{\sin A}{\cos A}=\frac{\frac{3}{5}}{\frac{5}{5}}=\frac{3}{4}$

Hence, the required value of $\tan A$ is $3 / 4$.

 

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