Question

Please give me the answer for the question.​

Answer 1

Answer:

Please find below the solution to the asked query

Step-by-step explanation:

Taking sin2A+sin2B-sin2C

= A + B + C = π

sin 2A + sin 2B - sin 2C = 4 cos A cos B sin C

From the double angle formula:

sin 2Θ = 2 sin Θ cos Θ

sin 2A + sin 2B - sin 2C

... = 2 sin A cos A + 2 sin B cos B - 2 sin C cos C

Since A + B + C = π;

A is a supplement angle of ( B + C )

B is a supplement angle of ( A + C )

C is a supplement angle of ( A + B )

TAKE NOTE that the sine of supplementary angles are equal !!!

sin 2A + sin 2B - sin 2C

... = 2 sin A cos A + 2 sin B cos B - 2 sin C cos C

... = 2 sin ( B + C ) cos A + 2 sin ( A + C ) cos B - 2 sin ( A + B ) cos C

From the Sum of Angle Identity:

sin ( α + ß ) = sin α cos ß + cos α sin ß

sin 2A + sin 2B - sin 2C

= 2 sin A cos A + 2 sin B cos B - 2 sin C cos C

= 2 sin ( B + C ) cos A + 2 sin ( A + C ) cos B - 2 sin ( A + B ) cos C

= 2 ( sin B cos C + cos B sin C ) cos A

+ 2 ( sin A cos C + cos A sin C ) cos B

– 2 ( sin A cos B + cos A sin B ) cos C

= 2 cos A sin B cos C + 2 cos A cos B sin C

+ 2 sin A cos B cos C + 2 cos A cos B sin C

– 2 sin A cos B cos C – 2 cos A sin B cos C

= 2 cos A cos B sin C + 2 cos A cos B sin C

= 4 cos A cos B sin C______(1)

Now taking sin2A+sin2B +sin2C

= If A+B+C=180 degree, then A+B=180-C hence sin(A+B) = sinC, cos(A+B)=-cosC

sin2A + sin 2B +sin2C

= 2 sin(A+B)cos(A-B) + 2sinC cosC

=2sinC cos(A-B)+2sinC cosC

=2sinC (cos(A-b) + cos C)

=2sin C(cos(A-B) - cos(A+B) )

= 2sinC . 2sin A sin B

=4 sinA sin B sin C______(2)

Now dividing (1)/(2) we get ,

sin2A + sin 2B +sin2C/sin2A+sin2B-sin2C=cotA cotB

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Hope this information will clear your doubts about this topic.(◕ᴗ◕✿)