If $$sin x =\dfrac{ 3}{5}, tan y = \dfrac{1}{2}$$and $$\dfrac{ π}{2} < x< π< y< \dfrac{3π}{2}$$ find the value of 8 tan x -$$\sqrt{5}$$ sec y.

Asked by Sakshi | 1 year ago |  80

#### 1 Answer

##### Solution :-

We know that, x is in second quadrant and y is in third quadrant.

In second quadrant, cos x and tan x are negative.

In third quadrant, sec y is negative.

By using the formula,

cos x = $$– \sqrt{(1-sin^2 x)}$$

tan x =$$\dfrac{ sin x}{cos x}$$

Now,

cos x = $$– \sqrt{(1-sin^2 x)}$$

= $$\dfrac{ – 4}{5}$$

tan x = $$\dfrac{ sin x}{cos x}$$

= $$\dfrac{3}{5}× \dfrac{-5}{4}$$

$$\dfrac{-3}{4}$$

We know that $$sec y = – \sqrt{(1+tan^2 y)}$$

$$\sqrt{-\dfrac{5}{2}}$$

= $$\dfrac{ (-12+5)}{2}$$

=$$\dfrac{-7}{2}$$

$$8 tan x – \sqrt{5} sec y = \dfrac{-7}{2}$$

Answered by Sakshi | 1 year ago

### Related Questions

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

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

Class 11 Maths Trigonometric Functions View Answer

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

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

Class 11 Maths Trigonometric Functions View Answer

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

Class 11 Maths Trigonometric Functions View Answer

#### prove that cos 570° sin 510° + sin (-330°) cos (-390°) = 0

prove that cos 570° sin 510° + sin (-330°) cos (-390°) = 0

Class 11 Maths Trigonometric Functions View Answer

#### prove that tan (-125°) cot (-405°) – tan (-765°) cot (675°) = 0

prove that tan (-125°) cot (-405°) – tan (-765°) cot (675°) = 0

Class 11 Maths Trigonometric Functions View Answer