Question

In Right triangle ABC right angle at B, If cos A= 4/5, then tan A =?
(a) 3/5
(b) 3/4
(c) 4/3
(d) 4/5

Specify with method.

Answer 1

Answer:

3/b (option b)

Step-by-step explanation:

Given that cos A = 4/5

As $sin^2A + vos^2A = 1$,

so, sin A = $\sqrt{ 1- cos^2A}$

So,

sin A

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

Now,

tan A

= sinA/cosA

= (3/5)/(4/5)

= (3/5) × (5/4)

= 3/4

Answer 2

Answer:

The correct answer is option (b)

Step-by-step explanation:

Given $\cos A=\frac{4}{5}$

$\Rightarrow \sec A=\frac{1}{\cos A}=\frac{5}{4}$

Also we know $\sec^2A-\tan^2A=1$

$\Rightarrow \frac{25}{16}-\tan^2A=1\\\Rightarrow \tan^2A=\frac{25}{16}-1=\frac{25-16}{16}=\frac{9}{16}\\\Rightarrow \tan A=\sqrt{\frac{9}{16}}=\frac{3}{4}$

Thus the answer is option (b)