Question

Solve it!!

$\Huge{Prove}[\tex]\ \textless \ br /\ \textgreater \ \ \textless \ br /\ \textgreater \ [tex] \frac{ \sin(A) - \sin( B) }{ \cos(A) + \cos(B) } + \frac{ \cos(A) - \cos (B) }{ \sin (A) + \sin (B) } = 0$


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Answer 1

Question:-

Prove that,

$\longrightarrow\sf{\dfrac {\sin A-\sin B}{\cos A+\cos B}+\dfrac {\cos A-\cos B}{\sin A+\sin B}=0}$

Solution:-

Assume the given equation to be true. Then,

$\longrightarrow\sf{\dfrac {\sin A-\sin B}{\cos A+\cos B}=-\dfrac {\cos A-\cos B}{\sin A+\sin B}\quad\quad\dots(1)}$

But remember the sum - to - product identities.

• $\sf{\sin A+\sin B=2\sin\left (\dfrac {A+B}{2}\right)\cos\left (\dfrac{A-B}{2}\right)}$

• $\sf{\sin A-\sin B=2\cos\left (\dfrac {A+B}{2}\right)\sin\left (\dfrac {A-B}{2}\right)}$

• $\sf{\cos A+\cos B=2\cos\left (\dfrac {A+B}{2}\right)}\cos\left(\dfrac {A-B}{2}\right)$

• $\sf{\cos A-\cos B=-2\sin\left (\dfrac {A+B}{2}\right)\sin\left (\dfrac {A-B}{2}\right)}$

Then (1) becomes,

$\longrightarrow\sf{\dfrac {2\cos\left (\dfrac {A+B}{2}\right)\sin\left (\dfrac {A-B}{2}\right)}{2\cos\left (\dfrac {A+B}{2}\right)\cos\left(\dfrac {A-B}{2}\right)}=-\dfrac {-2\sin\left (\dfrac {A+B}{2}\right)\sin\left (\dfrac {A-B}{2}\right)}{2\sin\left (\dfrac {A+B}{2}\right)\cos\left (\dfrac{A-B}{2}\right)}}$

From this we nicely get,

$\longrightarrow\sf{1=1}$

So the assumed equation is true. Hence the Proof!