Chapter 4, Problem 50CR

To determine

To Find: The value of all the trigonometric ratios in terms of $\theta$ given in the question.

Answer to Problem 50CR

  $\sin \theta =\frac{40}{41}$ , $\cos \theta =\frac{9}{41}$ , $\cos ec\theta =\frac{41}{40}$ , $\sec \theta =\frac{41}{9}$ , $\tan \theta =\frac{40}{9}$

Explanation of Solution

Given information: It is given that $\cot \theta =\frac{9}{40}$

Formula Used: Pythagoras theorem states that

  ${(Hypotenuse)}^2={(opposite)}^2+{(adjacent)}^2$

  $\sin \theta$ = $\frac{opposite}{hypotenuse}$

  $\cos \theta =\frac{adjacent}{hypotenuse}$

  $\tan \theta =\frac{opposite}{adjacent}$

  $\cos ec\theta =\frac{1}{\sin \theta }$

  $sec\theta =\frac{1}{\cos \theta }$

  $\cot \theta =\frac{1}{\tan \theta }$

According to the question

  $\cot \theta =\frac{9}{40}$

That means, $\frac{adjacent}{opposite}$ = $\frac{9}{40}$

Now, let adjacent = 9k and opposite = 40k

Now, by Pythagoras theorem

  ${(hypotenuse)}^2={(9k)}^2+{(40k)}^2$

  $hypotenuse=41k$

Now, since $\sin \theta =\frac{opposite}{hypotenuse}$

So, $\sin \theta =\frac{40k}{41k}$

Or $\sin \theta =\frac{40}{41}$

  $\cos \theta =\frac{adjacent}{hypotenuse}$

So, $\cos \theta =\frac{9k}{41k}$

Or $\cos \theta =\frac{9}{41}$

Similarly, $\cos ec\theta =\frac{41}{40}$ , $\sec \theta =\frac{41}{9}$ , $\tan \theta =\frac{40}{9}$