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

sin \frac{A}{2} . cos \frac{B}{2}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

sec \frac{A}{2} . c o s e c \frac{B}{2}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

tan \frac{A}{2} . cot \frac{B}{2}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

2 sin \frac{A}{2} . cos \frac{B}{2}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is C $tan \frac{A}{2} . cot \frac{B}{2}$
$\frac{1 - cos A + cos B - cos (A + B)}{1 + cos A - cos B - cos (A + B)} = \frac{2 {sin}^{2} \frac{A}{2} + 2 sin \frac{A}{2} sin \frac{A + 2 B}{2}}{2 {cos}^{2} \frac{A}{2} - 2 cos \frac{A + 2 B}{2} cos \frac{A}{2}}$

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

$= tan \frac{A}{2} \times \frac{2 sin \frac{A + B}{2} cos \frac{B}{2}}{2 sin \frac{B}{2} sin \frac{A + B}{2}}$

$= tan \frac{A}{2} cot \frac{B}{2}$

Suggest Corrections

Similar questions

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

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) $\frac{c o s A + c o s B}{c o s B - c o s A} = c o t \frac{A + B}{2} c o t \frac{A - B}{2}$