In ABC- sin A-B-sin A-B-MP PET 1986- A- a2-b2-c2B- a2-b2-c22C- c-a2-b2D- c2-a2-b2

The correct option is A a^2 b^2/c^2sin(A B)/sin(A+B = sin A cos B sin B cos A/sin C = a/c cos B b/c cos A But cos B = a^2 + c^2 b^2/2ac, cos A = b^2 + ...

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Standard XI

Mathematics

L'Hospital Rule to Remove Indeterminate Form

In ABC, sin A...

Question

In $△ A B C, \frac{s i n (A - B)}{s i n (A + B)} =$
[MP PET 1986]

\frac{a^{2} - b^{2}}{c^{2}}

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\frac{a^{2} + b^{2}}{c^{2}}

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\frac{c^{2}}{a^{2} - b^{2}}

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\frac{c^{2}}{a^{2} + b^{2}}

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Solution

The correct option is A $\frac{a^{2} - b^{2}}{c^{2}}$
$\frac{s i n (A - B)}{s i n (A + B} = \frac{s i n A c o s B - s i n B c o s A}{s i n C}$
$= \frac{a}{c} c o s B - \frac{b}{c} c o s A$
But cos $B = \frac{a^{2} + c^{2} - b^{2}}{2 a c}, c o s A = \frac{b^{2} + c^{2} - a^{2}}{2 b c}$
$\Rightarrow \frac{a}{c} c o s B - \frac{b}{c} c o s A = \frac{1}{2 c^{2}}$
$(a^{2} + c^{2} - b^{2} - b^{2} - c^{2} + a^{2}) = \frac{a^{2} - b^{2}}{c^{2}}$

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