If sinx - -sinx - - - ab- cosx - -cosx - - - AB and aB - bA - 0- then prove that - cos- - - - aA - bBaB - bA-

Adding the given relation , we get #160; ab + AB = sin(x α #160; + x β )sin(x β)cos (x β ) aB + bA bB = sin(2x ...

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Expansions to Remove Indeterminate Form

If sinx - ...

Question

If $\frac{s i n (x - α)}{s i n (x - β)} = \frac{a}{b}, \frac{c o s (x - α)}{c o s (x - β)} = \frac{A}{B}$ and aB + bA $\neq$ 0, then prove that -
$c o s (α - β) = \frac{a A + b B}{a B + b A} .$

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