If in any triangle ABC- a 2 - b 2 - a 2 - b 2 -sin A-B -sin A-B - then the triangle is

The correct option is D Right angled or isosceles #160; a ^ 2 b ^ 2 / a ^ 2 + b ^ 2 =sin( A B ) #160; #160;/sin( A+B ) #160; #160; Using comp ...

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Byju's Answer

Standard XII

Mathematics

Sum of Trigonometric Ratios in Terms of Their Product

If in any tri...

Question

If in any triangle $A B C; \frac{a^{2} - b^{2}}{a^{2} + b^{2}} = \frac{sin (A - B)}{sin (A + B)}$ , then the triangle is

Right angled

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Isosceles

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Isosceles right angled

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Right angled or isosceles

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Solution

The correct option is D Right angled or isosceles
$\frac{a^{2} - b^{2}}{a^{2} + b^{2}} = \frac{sin (A - B)}{sin (A + B)}$ Using componendo and dividendo

$\frac{2 a^{2}}{- 2 b^{2}} = \frac{2 sin A cos B}{- 2 cos A sin B} \Rightarrow \frac{a^{2}}{b^{2}} = \frac{sin A cos B}{cos A sin B}$

$\Rightarrow \frac{a}{b} = \frac{cos B}{cos A} \Rightarrow \frac{sin A}{sin B} = \frac{cos B}{cos A} \Rightarrow sin 2 A = sin 2 B$

$\Rightarrow 2 A = n π + {(- 1)}^{n} 2 B; n \in Z$

$\Rightarrow 2 A = 2 B 2 A + 2 B = π$

$\Rightarrow Δ$ is isosceles or right angled.

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Similar questions

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If in any triangle ABC;a2−b2a2+b2=sin(A−B)sin(A+B), then the triangle is

The correct option is D Right angled or isoscelesa2−b2a2+b2=sin(A−B)sin(A+B) Using componendo and dividendo2a2−2b2=2sinAcosB−2cosAsinB⇒a2b2=sinAcosBcosAsinB⇒ab=cosBcosA⇒sinAsinB=cosBcosA⇒sin2A=sin2B⇒2A=nπ+(−1)n2B;n∈Z⇒2A=2B2A+2B=π⇒Δ is isosceles or right angled.

If in any triangle $A B C; \frac{a^{2} - b^{2}}{a^{2} + b^{2}} = \frac{sin (A - B)}{sin (A + B)}$ , then the triangle is