In a triangle ABC , ∠ C = 9 0 ∘ , then a 2 + b 2 a 2 − b 2 is equal to :

A + B = 18 0 ∘ − C = 9 0 ∘ a = 2 R sin A , b = 2 R sin B , c = 2 R sin C ∴ a 2 + b 2 a 2 − b 2 = s i n 2 A + s i n 2 B s i n 2 A − s i n 2 B = s i n 2 A + s i n 2 ( 9 0 ∘ − A ) s i n ( A + B ) s i n ( A − B ) [ ∵ A + B = 9 0 ∘ ] = s i n 2 A + c o s 2 A s i n 9 0 ∘ s i n ( A − B ) = sin ( A − B )

Mathematics - Heights and Distances Question with Solution | TestHub

MathematicsHeights and DistancesGeneralMedium2 minPYQ_2009

MathematicsMediumsingle choice

In a triangle $ABC, ∠ C = 9 0^{\circ}$ , then $\frac{a ^{2} - b ^{2}}{a ^{2} + b ^{2}}$ is equal to :

Options:

Answer:

Solution:

$A + B = 18 0^{\circ} - C = 9 0^{\circ} a = 2 R sin A, b = 2 R sin B, c = 2 R sin C ∴ \frac{a ^{2} - b ^{2}}{a ^{2} + b ^{2}} = \frac{s i n ^{2} A - s i n ^{2} B}{s i n ^{2} A + s i n ^{2} B} = \frac{s i n ( A + B ) s i n ( A - B )}{s i n ^{2} A + s i n ^{2} ( 9 0 ^{\circ} - A )} [∵ A + B = 9 0^{\circ}] = \frac{s i n 9 0 ^{\circ} s i n ( A - B )}{s i n ^{2} A + c o s ^{2} A} = sin (A - B)$

Stream:BITSATSubject:MathematicsTopic:Heights and DistancesSubtopic:General

⏱ 2mℹ️ Source: PYQ_2009

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