Find the values of other five trigonometric functions if cot x = $$\dfrac{3}{4}$$, x lies in third quadrant.

Asked by Pragya Singh | 1 year ago |  61

##### Solution :-

It is given that

cot x =$$\dfrac{3}{4}$$

We can write it as

$$tanx= \dfrac{1}{cotx}=\dfrac{1}{\dfrac{3}{4}}=\dfrac{4}{3}$$

We know that

1 + tan2 x = sec2 x

We can write it as

1 + $$( \dfrac{4}{3})^2$$= sec2 x

Substituting the values

1 + $$\dfrac{16}{9}$$ = sec2 x

cos2 x = $$\dfrac{25}{9}$$

sec x = ± $$\dfrac{5}{3}$$

Here x lies in the third quadrant so the value of sec x will be negative

sec x = $$- \dfrac{5}{3}$$

We can write it as

Therefore,
$$cos x=\dfrac{1}{secx}$$

$$\dfrac{1}{(-\dfrac{5}{3})}$$

$$-\dfrac{3}{5}$$

Now, $$tanx=\dfrac{sinx}{cosx}$$

Therefore, sinx=tanxcosx

Hence we have, sinx = $$\dfrac{4}{3}\times (-\dfrac{3}{5})$$

$$(- \dfrac{4}{5})$$

And

$$cosecx=\dfrac{1}{sinx}$$

$$- \dfrac{5}{4}$$

Answered by Abhisek | 1 year ago

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