If A, B and C are interior angles of triangle ABC, then show that cosec²(b/2+c/2) - tan A/2 =1
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The question is wrong it should be tan^2 A/2 instead of tan A/2
LHS
Cosec^2 (b/2+c/2)-tan^2(a/2)
Now
b+c=180-a
Therefore
Cosec^2(180/2-a/2)
=Cosec^2(90-a/2)
=Sec^2(a/2)
LHS
Sec^2(a/2)-tan^2 (a/2)
=1
=RHS
Hence proved
LHS
Cosec^2 (b/2+c/2)-tan^2(a/2)
Now
b+c=180-a
Therefore
Cosec^2(180/2-a/2)
=Cosec^2(90-a/2)
=Sec^2(a/2)
LHS
Sec^2(a/2)-tan^2 (a/2)
=1
=RHS
Hence proved
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