Find the values of other five trigonometric functions if sin x = $$\dfrac{3}{5}$$, x lies in second quadrant.

Asked by Pragya Singh | 1 year ago |  61

#### 1 Answer

##### Solution :-

It is given that

sin x = $$\dfrac{3}{5}$$

We can write it as

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

$$\dfrac{1}{\dfrac{3}{5}}$$$$\dfrac{5}{3}$$

We know that

sin2 x + cos2 x = 1

We can write it as

cos2 x = 1 – sin2 x

Substituting sin x =$$\dfrac{3}{5}$$  in the formula, we obtain,

$$cos^2x=1- (\dfrac{3}{5})^2$$

$$cos^2x=1- (\dfrac{9}{25})$$ = $$\dfrac{16}{25}$$

cosx= $$\pm \dfrac{4}{5}$$

Since x lies in the 2nd quadrant, the value of cos x will be negative

cosx = $$-\dfrac{4}{5}$$

Therefore, secx = $$\dfrac{1}{cosx}$$

$$\dfrac{1}{(-\dfrac{4}{5})}=- \dfrac{5}{4}$$

Hence,

$$tanx=\dfrac{sinx}{cosx}$$

$$\dfrac{(\dfrac{3}{5})}{(-\dfrac{4}{5})}=\dfrac{1}{2}=-\dfrac{3}{4}$$

And

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

Answered by Abhisek | 1 year ago

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