If sin A -sin B - and cos A -cos B - then write the value of tan A - B -2

We have, sinA+sinB=α and cosA+cosB=β …(ii) sinA+sinB/cosA+cosB=α/β ⇒2sin ( A+B/2 )cos ( A B/2 )/2cos ( A+B/2 )cos ( A B/2 )=α/β ⇒sin ( A+B/2 )/cos ( A+B/2 )= ...

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Byju's Answer

Standard XII

Mathematics

Trigonometric Equations

If sin A +sin...

Question

If $s i n A + s i n B = α$ and $c o s A + c o s B = β$ , then write the value of $t a n (\frac{A + B}{2})$

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Solution

We have,
$s i n A + s i n B = α$
and $c o s A + c o s B = β \dots (i i)$
$\frac{s i n A + s i n B}{c o s A + c o s B} = \frac{α}{β} \Rightarrow \frac{2 s i n (\frac{A + B}{2}) c o s (\frac{A - B}{2})}{2 c o s (\frac{A + B}{2}) c o s (\frac{A - B}{2})} = \frac{α}{β} \Rightarrow \frac{s i n (\frac{A + B}{2})}{c o s (\frac{A + B}{2})} = \frac{α}{β} \Rightarrow t a n (\frac{A + B}{2}) = \frac{α}{β}$

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