Find the values of other five trigonometric functions if $$tan x=-\dfrac{5}{12}$$, x lies in second quadrant.

Asked by Pragya Singh | 1 year ago |  65

##### Solution :-

Here given that, $$tanx=-\dfrac{5}{12}$$

Therefore we have,

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

$$\dfrac{1}{(-\dfrac{5}{12})}= -\dfrac{12}{5}$$

We know that

sec2 x - tan2 x = 1

We can write it as

sec2 x = 1 + tan2 x

Substituting tanx = $$- \dfrac{5}{12}$$ in the formula, we obtain,

$$sec^2x=1+(-\dfrac{5}{12})^2$$

$$sec^2x=1+\dfrac{25}{144}$$

$$\dfrac{169}{144}$$

$$secx\pm \dfrac{13}{12}$$

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

$$secx=- \dfrac{13}{12}$$

Therefore, $$cosx=\dfrac{1}{secx}$$

$$- \dfrac{12}{13}$$

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

Therefore, sinx=tanxcosx

Hence we have,$$sinx=(-\dfrac{5}{12})\times - \dfrac{12}{13}$$

$$=( \dfrac{5}{13})$$

$$cosecx=\dfrac{1}{sinx}= \dfrac{13}{5}$$

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