Question

sinx+siny=a and cosx-cosy=b then tan(x-y)/2=​

Answer 1

Answer:

Step-by-step explanation:

We have,

$\tt{sin(x)+sin(y)=a\,\,\,\,and\,\,\,\,cos(x)-cos(y)=b}$

Now,

$\sf{sin(x)+sin(y)=a}$

$\sf{\implies\,2sin\bigg(\dfrac{x+y}{2}\bigg)cos\bigg(\dfrac{x-y}{2}\bigg)=a\,\,\,\,....(1)}$

Also,

$\sf{cos(x)-cos(y)=b}$

$\sf{\implies\,-2sin\bigg(\dfrac{x+y}{2}\bigg)sin\bigg(\dfrac{x-y}{2}\bigg)=b\,\,\,\,....(2)}$

Divide (1) by (2),

$\sf{\implies\,\dfrac{2sin\bigg(\dfrac{x+y}{2}\bigg)cos\bigg(\dfrac{x-y}{2}\bigg)}{-2sin\bigg(\dfrac{x+y}{2}\bigg)sin\bigg(\dfrac{x-y}{2}\bigg)}=\dfrac{a}{b}}$

$\sf{\implies\,-\dfrac{cos\bigg(\dfrac{x-y}{2}\bigg)}{sin\bigg(\dfrac{x-y}{2}\bigg)}=\dfrac{a}{b}}$

$\sf{\implies\,-cot\bigg(\dfrac{x-y}{2}\bigg)=\dfrac{a}{b}}$

$\sf{\implies\,cot\bigg(\dfrac{x-y}{2}\bigg)=-\dfrac{a}{b}}$

$\sf{\implies\,tan\bigg(\dfrac{x-y}{2}\bigg)=-\dfrac{b}{a}}$

Answer 2

Answer:

$\mathrm{\dfrac{-b}{a}}$

Step-by-step explanation:

Given :

$\bullet \ \mathrm{sinx + siny = a}$

$\bullet \ \mathrm{cosx - cosy = b}$

To find :

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

Solution :

We know that,

$\begin{gathered}\boxed{\begin{aligned} \mathrm { \bullet \ sinC + sinD} = & \mathrm{2sin\bigg(\dfrac{C+D}{2}\bigg)cos\bigg(\dfrac{C-D}{2}\bigg)} \\\frac{\qquad \qquad \qquad \qquad }{} & \frac{\qquad \qquad \qquad \qquad\qquad\qquad }{} \\ \mathrm { \bullet \ cosC - cosD} = & \mathrm{-2sin\bigg(\dfrac{C+D}{2}\bigg)sin\bigg(\dfrac{C-D}{2}\bigg)} \end{aligned}} \end{gathered}$

As here,

$\longmapsto \mathrm{sinx + siny = a} \ \text{and} \ \mathrm{cosx - cosy = b}$

$\implies \mathrm {sinx + siny = 2sin\bigg(\dfrac{x+y}{2}\bigg)cos\bigg(\dfrac{x-y}{2}\bigg) = a}\ \ \ ---- [1]$

$\implies \mathrm {cosx - cosy = -2sin\bigg(\dfrac{x+y}{2}\bigg)sin\bigg(\dfrac{x-y}{2}\bigg) = b} \ \ \ - - - - [2]$

Dividing 1 by 2, we get,

$\implies \mathrm{\dfrac{2sin\bigg(\dfrac{x+y}{2}\bigg)cos\bigg(\dfrac{x-y}{2}\bigg)}{-2sin\bigg(\dfrac{x+y}{2}\bigg)sin\bigg(\dfrac{x-y}{2}\bigg)} = \dfrac{a}{b}}$

$\implies \mathrm{\dfrac{\not2\bcancel{sin\bigg(\dfrac{x+y}{2}\bigg)}{cos\bigg(\dfrac{x-y}{2}\bigg)}}{\not2\bcancel{sin\bigg(\dfrac{x+y}{2}\bigg)}sin\bigg(\dfrac{x-y}{2}\bigg)} = \dfrac{-a}{b}}$

$\implies \mathrm{cot\bigg(\dfrac{x-y}{2}\bigg) = \dfrac{-a}{b}}$

We know that,

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

So,

$\therefore \underline{\boxed{\mathbf{tan\bigg(\dfrac{x-y}{2}\bigg) = \dfrac{-b}{a}}}}$

Hope it helps!!