Prove that - sin A - sin Bcos A - cos B - cos A - cos Bsin A - sin B - 0

L.H.S. = sin A sin Bcos A + cos B + cos A cos Bsin A + sin B ⇒ L.H.S. = (sin A sin B) (si ...

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Basics of Geometry

Prove that : ...

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Prove that :
$\frac{sin A - sin B}{cos A + cos B} + \frac{cos A - cos B}{sin A + sin B} = 0$

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$\frac{s i n A - s i n B}{c o s A + c o s B} + \frac{c o s A - c o s B}{s i n A + s i n B} =$

Simplify the expression:
$\frac{s i n A - s i n B}{c o s A + c o s B} + \frac{c o s A - c o s B}{s i n A + s i n B}$

{(\frac{c o s A + c o s B}{s i n A - s i n B})}^{2014} + {(\frac{s i n A + s i n B}{c o s A - c o s B})}^{2014} = . . . . . . . . . . .

{(\frac{cos A + cos B}{sin A - sin B})}^{n} + {(\frac{sin A + sin B}{cos A - cos B})}^{n}

= 2 {cot}^{n} (\frac{A - B}{2})

n

= 0

n

\frac{cos A + cos B}{sin A - sin B} = cot \frac{A - B}{2}

Show that: $\frac{(sin A - sin B)}{(cos A + cos B)} + \frac{(cos A - cos B)}{(sin A + sin B)} = 0$

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