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

##### 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

### Related Questions

#### prove that sin 8π/3 cos 23π/6 + cos 13π/3 sin 35π/6 = 1/2

prove that $$sin \dfrac{8π}{3} cos \dfrac{23π}{6} + cos \dfrac{13π}{3} sin \dfrac{35π}{6} = \dfrac{1}{2}$$

#### prove that 3 sin π/6 sec π/3 – 4 sin 5π/6 cot π/4 = 1

prove that $$3 sin \dfrac{π}{6} sec \dfrac{π}{3} – 4 sin \dfrac{5π}{6} cot \dfrac{π}{4} = 1$$

#### prove that tan 11π/3 – 2 sin 4π/6 – 3/4 cosec2 π/4 + 4 cos2 17π/6 = (3 – 4\sqrt{3})/2

prove that $$tan \dfrac{11π}{3} – 2 sin \dfrac{4π}{6} – \dfrac{3}{4} cosec^2 \dfrac{π}{4} + 4 cos^2 \dfrac{17π}{6} = \dfrac{(3 – 4\sqrt{3})}{2}$$