71. 1-cosA+cosB-cos(A+B)/1+cosA-cosB-cos(A+B) equals to what

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\frac{1 - cos A + cos B - cos (A + B)}{1 + cos A - cos B - cos (A + B)} =

Prove that $\frac{1 - cos A + cos B - cos (A + B)}{1 + cos A - cos B - cos (A + B)} = tan \frac{A}{2} cot (\frac{B}{2})$ .

A + B = \frac{π}{3}; cos A + cos B = 1

, value of

| cos A - cos B |

Prove that:
(i) $\frac{s i n A + s i n 3 A}{c o s A - c o s 3 A} = c o t A$
(ii) $\frac{s i n 9 A - s i n 7 A}{c o s 7 A - c o s 9 A} = c o t 8 A$
(iii) $\frac{s i n A - s i n B}{c o s A - c o s B} = t a n \frac{A - B}{2}$
(iv) $\frac{s i n A + s i n B}{s i n A - s i n B} t a n (\frac{A + B}{2}) c o t \frac{A + B}{2}$
(v) $\frac{c o s A + c o s B}{c o s B - c o s A} = c o t \frac{A + B}{2} c o t \frac{A - B}{2}$

Q. If

A + B = \frac{π}{3}

and

cos A + cos B = 1

then

cos (A - B)

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