Prove that $$\dfrac{sinx-siny}{cosx+cosy}=tan\dfrac{x-y}{2}$$

Asked by Pragya Singh | 1 year ago |  51

##### Solution :-

We know that,

sinA-sinB=2cos$$( \dfrac{A+B}{2})sin( \dfrac{A-B}{2})$$,

cosA+cosB=2cos$$( \dfrac{A+B}{2})cos( \dfrac{A-B}{2})$$

$$\dfrac{sinx-siny}{cosx+cosy}=tan\dfrac{x-y}{2}$$

L.H.S.$$\dfrac{sinx-siny}{cosx+cosy}$$

$$\dfrac{2cos(\dfrac{x+y}{2}).sin(\dfrac{x-y}{2})}{2cos(\dfrac{x+y}{2}).cos(\dfrac{x-y}{2})}$$

$$\dfrac{sin(\dfrac{x-y}{2})}{cos(\dfrac{x-y}{2})}$$

$$tan(\dfrac{x-y}{2})$$

L.H.S=R.H.S

Hence proved.

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

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

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

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

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