Match the entries of Column-I and Column-II.
$cos A + cos B = a$ , $sin A + sin B = b$

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Solution

(a) $\frac{b}{a} = \frac{s i n A + s i n B}{c o s A + c o s B}$
$= \frac{2 s i n \frac{A + B}{2} c o s \frac{A - B}{2}}{2 c o s \frac{A + B}{2} c o s \frac{A - B}{2}} = \frac{A + B}{2} = T$ say
$∴$ $(a) \to (q)$
(b) $c o s (A + B) = \frac{1 - T^{2}}{1 + T^{2}} = \frac{a^{2} - b^{2}}{a^{2} + b^{2}}$
$∴$ $(b) \to (r)$
(c) $a^{2} + b^{2} = {(c o s A + c o s B)}^{2} + {(s i n A + s i n B)}^{2}$
$= ({c o s}^{2} A + {s i n}^{2} A) + ({s i n}^{2} B + {s i n}^{2} B) + 2 (c o s A c o s B + s i n A s i n B)$
$∴$ $(a^{2} + b^{2} - 2) = 2 c o s (A - B)$
$∴$ $(c) \to (s)$
(d) $s i n (A + B) = \frac{2 T}{1 + T^{2}} = \frac{2 a b}{a^{2} + b^{2}}$
$∴$ $(d) \to (p)$

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