Question

If tan A=√3, find the other trigonometric ratio of angle A.

Answer 1

Step-by-step explanation:

if

$\tan(a) = \sqrt{3} \\ then \: the \: a \: = \: {60}^{0} \\ \sin(a) = \frac{ \sqrt{3} }{2} \\ \cos(a ) = \frac{1}{2} \\ \cot(a) = \frac{1}{ \sqrt{3} } \\ \sec(a) = 2 \\ \csc(a) = \frac{2}{ \sqrt{3} }$

Answer 2

$\bf{\underline{Solution}}:-$

$\rm\:tanA=\dfrac{p}{b}=\dfrac{BC}{AB}=\dfrac{\sqrt{3}}{1}$

Let BC be √3 k

and AB be k

Now,

In right ∆ ABC

$\rm\:AC=\sqrt{{(AB)}^{2}+{(BC)}^{2}}$

$\rm\longrightarrow\:AC=\sqrt{{(k)}^{2}+{(\sqrt{3})^2}}$

$\rm\longrightarrow\:AC=\sqrt{{k}^{2}+{3k}^{2}}$

$\rm\longrightarrow\:AC=\sqrt{{4k}^{2}}$

$\rm\longrightarrow\:AC=2k$

Now,

$\rm\therefore\:sinA=\dfrac{p}{h}=\dfrac{BC}{AC}$

$\rm\longrightarrow\:sinA=\dfrac{\sqrt{3}\cancel{k}}{2\cancel{k}}$

$\rm\longrightarrow\:sinA=\dfrac{\sqrt{3}}{2}$

$\rm\therefore\:cosA=\dfrac{b}{h}=\dfrac{AB}{AC}$

$\rm\longrightarrow\:cosA=\dfrac{\cancel{k}}{2\cancel{k}}$

$\rm\longrightarrow\:cosA=\dfrac{1}{2}$

$\rm\therefore\:tanA=\dfrac{p}{b}=\dfrac{BC}{AB}$

$\rm\longrightarrow\:tanA=\dfrac{\sqrt{3}\cancel{k}}{\cancel{k}}$

$\rm\longrightarrow\:tanA={\sqrt{3}}$

$\rm\therefore\:cosecA=\dfrac{h}{p}=\dfrac{AC}{BC}$

$\rm\longrightarrow\:cosecA=\dfrac{2k}{\sqrt{3}k}$

$\rm\longrightarrow\:cosecA=\dfrac{2}{\sqrt{3}}$

$\rm\therefore\:secA=\dfrac{h}{b}=\dfrac{AC}{AB}$

$\rm\longrightarrow\:secA=\dfrac{2k}{k}$

$\rm\longrightarrow\:secA=2$

$\rm\therefore\:cotA=\dfrac{b}{p}=\dfrac{AB}{BC}$

$\rm\longrightarrow\:cotA=\dfrac{k}{\sqrt{3}k}$

$\rm\longrightarrow\:cotA=\dfrac{1}{\sqrt{3}}$