Question

If $\csc \theta =\sqrt{10}$ find the value of other trigonometrical ratios for $\theta$.

Answer 1

$\csc \theta =\sqrt{10}$

$\Rightarrow \sin \theta =\frac{1}{\csc \theta }=\frac{1}{\sqrt{10}}$ using reciprocal relations of trigonometry.

And we know that $\sin \theta =\frac{Opp.side}{hypotenuse}=\frac{1}{\sqrt{10}}$

$\therefore$Length of opposite side$=1$unit

and Length of hypotenuse$=\sqrt{10}$ units.

By Pythagoras theorem, we have

square of hypotenuse$=$sum of the squares of its base and the opposite sides$

Hence ${\left(\sqrt{10}\right)}^2={\left(base\right)}^2+1^2$

$\Rightarrow$Length of the base$=\sqrt{10-1}=\sqrt{9}=\pm 3$

Since $\csc \theta$ is positive in first and second quadrant.$\therefore$ base can be negative.

Hence base takes the values $3,-3$

$\therefore \sin \theta =\frac{1}{\sqrt{10}}$ in both first and second quadrant

$\cos \theta =\frac{3}{\sqrt{10}}$ in first quadrant and $\frac{-3}{\sqrt{10}}$ in second quadrant.

$\sec \theta =\frac{\sqrt{10}}{3}$ in first quadrant and $\frac{-\sqrt{10}}{3}$ in second quadrant.

$\tan \theta =\frac{1}{3}$ in first quadrant and $\frac{-1}{3}$ in second quadrant.

$\cot \theta =3$ in first quadrant and $\cot \theta =-3$ in second quadrant.