Question

if sin x + sin y = 4/5 and sin x - sin y = 2/7 prove that 5 tan (y + x)/2 + 14 tan( y - x)/2 = 0​

Answer 1

Answer:

Step-by-step explanation:

Given,

sinx + siny = 4/5 [Let it be equation - 1]

sinx - siny = 2/7 [Let it be equation - 2]

To Prove :-

$5tan\bigg(\dfrac{x+y}{2}\bigg)+14tan\bigg(\dfrac{y-x}{2}\bigg)=0$

Solution :-

Equation -1/Equation - 2

$\implies \dfrac{sinx+siny}{sinx-siny}=\dfrac{\dfrac{4}{5}}{\dfrac{2}{7}}$

$\dfrac{sinx+siny}{sinx-siny}=\dfrac{4}{5}\times \dfrac{7}{2}$

$\dfrac{sinx+siny}{sinx-siny}=\dfrac{14}{5}$

We know that :-

$sinC+sinD=2sin\bigg(\dfrac{C+D}{2}\bigg)cos\bigg(\dfrac{C-D}{2}\bigg)$

$sin-sinD=2cos\bigg(\dfrac{C+D}{2}\bigg)sin\bigg(\dfrac{C-D}{2}\bigg)$

$\implies \dfrac{2sin\bigg(\dfrac{x+y}{2}\bigg)cos\bigg(\dfrac{x-y}{2}\bigg)}{2cos\bigg(\dfrac{x+y}{2}\bigg)sin\bigg(\dfrac{x-y}{2}\bigg)}=\dfrac{14}{5}$

$tan\bigg(\dfrac{x+y}{2}\bigg)cot\bigg(\dfrac{x-y}{2}\bigg)=\dfrac{14}{5}$

$tan\bigg(\dfrac{x+y}{2}\bigg)\times\dfrac{1}{tan\bigg(\dfrac{x-y}{2}\bigg)}=\dfrac{14}{5}$

$5tan\bigg(\dfrac{x+y}{2}\bigg)=14tan\bigg(\dfrac{x-y}{2}\bigg)$

$5tan\bigg(\dfrac{x+y}{2}\bigg)-14tan\bigg(\dfrac{x-y}{2}\bigg)=0$

$5tan\bigg(\dfrac{x+y}{2}\bigg)-14tan\bigg(\dfrac{-(y-x)}{2}\bigg)=0$

$5tan\bigg(\dfrac{y+x}{2}\bigg)+14tan\bigg(\dfrac{y-x}{2}\bigg)=0$