Question

If sinA=1/3 then find all other trigonometric ratios

Answer 1

Given :-

$\sf sinA = \dfrac{1}{3}$

To Find :-

$\bullet \sf \ cosecA \\\\\bullet \ cosA \\\\\bullet \ secA \\\\\bullet \ tanA \\\\\bullet \ cotA$

Solution :-

We know :-

$\bf cosecA = \dfrac{1}{sinA}$

$\sf\therefore cosecA = 3$

We know :-

$\bf sin^2A+cos^2A = 1$

$\longrightarrow \sf \dfrac{1}{3}^2+cos^2A = 1 \\\\\longrightarrow cos^2A = 1- \dfrac{1}{9}\\\\\longrightarrow cos^2A = \dfrac{9-1}{9} \\\\\longrightarrow cos^2A = \dfrac{8}{9} \\\\\longrightarrow cosA = \sqrt{\dfrac{8}{9}} \\\\\longrightarrow \bf cosA = \dfrac{\sqrt{8 }}{3}$

We know :-

$\bf secA = \dfrac{1}{cosA}$

$\therefore \sf secA = \dfrac{3}{\sqrt{8} }$

We know :-

$\bf tanA = \dfrac{sinA}{cosA}$

$\therefore \sf tanA = \dfrac{1}{3} \times \dfrac{3}{\sqrt{8} }$

           $= \sf \dfrac{1}{\sqrt{8} }$

We know :-

$\bf cotA = \dfrac{1}{tanA}$

$\therefore \sf cotA = \sqrt{8}$

Answer 2

sinA = 1/3

from the above

opposite side = 1

hypotnuse = 3

adjacent side = x^2+1^2 = 3^2 ( pythogoras therom sum of squares of sides = hypotnuse square )

=x^2 +1 = 9

= x^2 = 8

= x = $\sqrt{8}$

= x = $\sqrt{4×2}$

x = $2\sqrt{2}$

so cosA = $\frac{adjacent side}{hypotnuse}$

= $\frac{2\sqrt{2}}{3}$

tanA = $\frac{sinA}{cosA}$

= $\frac{1}{2\sqrt{2}}$

CosecA = $\frac{1}{sinA}$

=$\frac{1}{\frac{1}{3}}$

= $3$

cotA = $\frac{1}{tanA}$

= $2\sqrt{2}$

secA = $\frac{1}{cosA}$

= $\frac{1}{2\sqrt{2}}$

= $2\sqrt{2}$