Question

4) If cos A = 4/5 then the value of tan A is a) 3/5 b) 3/4 c) 4/3 d) 5/3​

Answer 1

Answer:

sin A=√(1-cos^2 A)=3/5

tan A= sin A/cos A=(3/5)/(4/5)=3/4

Answer 2

Given :-

$\quad \leadsto \quad \sf \cos A = \dfrac{4}{5}$

To Find :-

The value of $\sf \tan A$

Solution :-

We Knows that ;

• $\sf \sin² \phi + \cos² \phi = 1$

Using this we have ;

$\quad \leadsto \quad \sf \sin² A + \cos² A = 1$

${ : \implies \quad \sf \sin² A + {\bigg( \dfrac{4}{5} \bigg)}^{2} = 1 }$

${ : \implies \quad \sf \sin²A + \dfrac{16}{25} = 1}$

${ : \implies \quad \sf \sin² A = 1 - \dfrac{16}{25}}$

${ : \implies \quad \sf \sin² A = \dfrac{25-16}{25}}$

${ : \implies \quad \sf \sin² A = \dfrac{9}{25}}$

${ : \implies \quad \sf ( \sin A)² = \pm {\bigg( \dfrac{3}{5} \bigg)}^{2}}$

${ : \implies \quad \sf ( \sin A)^{\cancel{2}}= \pm {\bigg( \dfrac{3}{5} \bigg)}^{\cancel{2}}}$

${ : \implies \quad \sf ( \sin A)= \pm {\bigg( \dfrac{3}{5} \bigg)}}$

But , Length can't be - ve

${ : \implies \quad \bf \therefore \quad \sin A = \dfrac{3}{5}}$

Now , we also knows that ;

• $\sf \tan \phi = \dfrac{\sin \phi}{\cos \phi}$

Using this we have ;

$\quad \leadsto \quad \sf \tan A = \dfrac{\sin A}{\cos A}$

${ : \implies \quad \sf \tan A = \dfrac{\dfrac{3}{5}}{\dfrac{4}{5}}}$

${ : \implies \quad \sf \tan A = \dfrac{\dfrac{3}{\cancel{5}}}{\dfrac{4}{\cancel{5}}}}$

${ : \implies \quad \bf \tan A = \dfrac{3}{4}}$

Henceforth , Option (b) $\bf \dfrac{3}{4}$ is correct :)