Math, asked by ritaalee21, 5 months ago

If A, B, C are the interior angles of a triangle ABC, then

sin (B+C)/2×cosA/2 + cos(B+C)/2×sin A/2

is equal to....

plz solve this in a paper..​

Answers

Answered by mathdude500
2

Answer:

we \: know \: in \: triangle \: abc \\ a + b + c = 180 \\ so \: b + c = 180 - a \\ so \:  \frac{b + c}{2}  =  \frac{180 - a}{2}  \\  =  >  \frac{b + c}{2}  = 90 -  \frac{a}{2}  \\  \\

sin\frac{b + c}{2} \: cos \frac{a}{2}  + cos\frac{b + c}{2} \: sin \frac{a}{2}  \\ =  sin (90 -  \frac{a}{2} )cos \frac{a}{2}  + cos(90 -  \frac{a}{2} )sin \frac{a}{2}   \\ =  cos \frac{a}{2} cos \frac{a}{2}  + sin \frac{a}{2} sin \frac{a}{2}  \\  =  {cos}^{2} \frac{a}{2}   +  {sin}^{2}  \frac{a}{2}  \\  = 1

Answered by suman8615
0

Answer:

this is correct................

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