Question

1) In AABC prove that
i) sin2A+ sin2B - sin2C = 4cos A cosB sin​

Answer 1

Answer:

Given:

A+B+C = 180°

=> A+B = 180° - C

=> sin(A+B) = sin(180-C)

= sinC -----(1)

______________________

We know that,

i))SinC - SinD = 2cos[(C+D)/2]sin[(C-D)/2]

ii) SinC + sinD = 2sin[(C+D)/2]cos[(C+D)/2]

iii) Sin2B = 2sinBcosB

_______________________

Here,

LHS = sin2A+sin2B-sin2C

= sin2A-sin2C+sin2B

=2cos[(2A+2C)/2]sin[(2A-2C)/2]+sin2B

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

= -2cosBsin(A-C)+2sinBcosB

= (2cosB)[-sin(A-C)-sinB]

= 2cosB[-sin(A-C)+sin(A+C)]

= 2cosB[sin(A+C)-sin(A-C)]

= 2cosB{2cos[(A+C+A-C)/2]sin[(A+C-A+C)/2]}

= 4cosBcosAsinC

= 4cosAcosBsinC

= RHS

Step-by-step explanation:

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Answer 2

$\huge\star\mathfrak\blue{{Answer:-}}$

Solution : L.H.S. = sin2A + sin2B - sin2c

$= 2sin( \frac{2a + 2b}{ 2} )cos( \frac{2a+ 2b}{2} ) - sin2c$

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

= 2sin (π-C) cos (A-B) - 2sinC cos [π - (A+R)

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

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

- 2 sinc.2cos (A-B+A+B/2) cos(A-B-A-B /2)

= 4 sinC cosA cosB

= 4 cosA cosB sinC

= R.H.S.