In each of the following, one of the six trigonometric ratios is given. Find the values of the other trigonometric ratios.$cosec\ \theta =\sqrt{10}$

Given:

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

To do:

We have to find the values of the other trigonometric ratios.

Solution:  

We know that,

In a right-angled triangle $ABC$ with right angle at $B$,

By Pythagoras theorem,

$AC^2=AB^2+BC^2$

By trigonometric ratios definitions,

$sin\ A=\frac{Opposite}{Hypotenuse}=\frac{BC}{AC}$

$cos\ A=\frac{Adjacent}{Hypotenuse}=\frac{AB}{AC}$

$tan\ A=\frac{Opposite}{Adjacent}=\frac{BC}{AB}$

$cosec\ A=\frac{Hypotenuse}{Opposite}=\frac{AC}{BC}$

$sec\ A=\frac{Hypotenuse}{Adjacent}=\frac{AC}{AB}$

$cot\ A=\frac{Adjacent}{Opposite}=\frac{AB}{BC}$

Here,

Let $cosec\ \theta=\frac{AC}{BC}=\frac{\sqrt{10}}{1}$

$AC^2=AB^2+BC^2$

$\Rightarrow (\sqrt{10})^2=(AB)^2+(1)^2$

$\Rightarrow AB^2=10-1$

$\Rightarrow AB=\sqrt{9}=3$

Therefore,

$sin\ \theta=\frac{BC}{AC}=\frac{1}{\sqrt{10}}$

$cos\ \theta=\frac{AB}{AC}=\frac{3}{\sqrt{10}}$

$tan\ \theta=\frac{BC}{AB}=\frac{1}{3}$

$sec\ \theta=\frac{AC}{AB}=\frac{\sqrt{10}}{3}$

$cot\ \theta=\frac{AB}{BC}=\frac{3}{1}=3$ 

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Updated on: 10-Oct-2022

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