Question

9) If cos A = 4/5, then tan A=?​

Answer 1

cos A = 4/5

=> tan A = 3/4

Answer 2

GiVen:-

• cosA = $\sf \dfrac{4}{5}$

To find:-

• Value of tanA = ?

SoluTion:-

GivEn that,

→ cosA = $\sf \dfrac{4}{5} = \dfrac{B}{H}$

Therefore,

Base (B) = 4 and Hypotenuse (H) = 5

★ UsiNg PyThagOraS ThEoreM :-

★ H² = P² + B² ★

$:\implies$ P² = H² - B²

$:\implies$ P² = (5)² - (4)²

$:\implies$ P² = 25 - 16

$:\implies$ P² = 9

$\small\sf\;\;\dag\;{\underline{Taking\;sqrt\;both\;sides}$

$:\implies$ $\sqrt{P^2} = \sqrt{9}$

$:\implies\bf \red{P = 3}$

Therefore,

★ tanA = $\sf \dfrac{P}{B} = \dfrac{3}{4}$

Similarly, we can find all trigonometric ratio from given one.

$\rule{150}{4}$

★ 2nd Method:-

cosA = $\sf \dfrac{4}{5}$

$\therefore$ secA = $\frac{1}{ \frac{4}{5}}$

$\implies$ secA = $\sf \dfrac{5}{4}$

As we know the identity:-

★ ${\underline{\boxed{\bf{\blue{ \sec^2{ \theta} = \tan^2{ \theta} + 1}}}}}$

Here, $\theta$ = A

$\implies\sf {\underline{ \sec^2{A} = \tan^2{A} + 1}}$

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

$\implies\sf \dfrac{25}{16} = \tan^2{A} + 1$

$\implies\sf \dfrac{25}{16} - 1 = \tan^2{A}$

$\implies\sf \dfrac{25 - 16}{16} = \tan^2{A}$

$\implies\sf \dfrac{9}{16} = \tan^2{A}$

$\small\sf\;\;\dag\;{\underline{Taking\;sqrt\;both\;sides}$

$\implies\sf \sqrt{ \dfrac{9}{16}} = \sqrt{ \tan^2{A}}$

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

$\implies{\underline{\boxed{\bf{\blue{ \tan{A} = \dfrac{3}{4}}}}}}$

$\dag\;{\underline{\underline{\bf{\pink{Hence\;SolvEd!!}}}}}$

$\rule{150}{4}$