If csc A -csc B -csc C -0 then show that sin A -sin B -sin C 2 -sin 2 A -

A + B + C =0 1 sin A #160; + 1 sin B #160; + 1 sin C #160; =0 sin B sin C +sin A sin B +sin A sin C #160; sin A sin B sin C ...

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Byju's Answer

Standard IX

Mathematics

Use of Trigonometric Table

If csc A +c...

Question

If $csc A + csc B + csc C = 0$ then show that ${(sin A + sin B + sin C)}^{2} = \sum {sin}^{2} A .$

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Solution

$csc A + csc B + csc C = 0 \frac{1}{sin A} + \frac{1}{sin B} + \frac{1}{sin C} = 0 \frac{sin B sin C + sin A sin B + sin A sin C}{sin A sin B sin C} = 0 sin B sin C + sin A sin B + sin A sin C = 0 Σ sin A sin B = 0$
Now L.H.S
${(sin A + sin B + sin C)}^{2}$
${sin}^{2} A + {sin}^{2} B + {sin}^{2} C + 2 \sum sin A sin B$
${sin}^{2} A + {sin}^{2} B + {sin}^{2} C \neq \sum sin A$ =R.H.S
LHS=RHS
Hence proof

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Similar questions

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If cscA+cscB+cscC=0 then show that (sinA+sinB+sinC)2=∑sin2A.

cscA+cscB+cscC=01sinA+1sinB+1sinC=0sinBsinC+sinAsinB+sinAsinCsinAsinBsinC=0sinBsinC+sinAsinB+sinAsinC=0ΣsinAsinB=0Now L.H.S(sinA+sinB+sinC)2sin2A+sin2B+sin2C+2∑sinAsinBsin2A+sin2B+sin2C≠∑sinA=R.H.SLHS=RHSHence proof

If $csc A + csc B + csc C = 0$ then show that ${(sin A + sin B + sin C)}^{2} = \sum {sin}^{2} A .$