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In usual notation show that (c^2-a^2+b^2)tan A=(a^2-b^2+c^2) tan B = (b^2-c^2+a^2)tanC

In usual notation show that (c^2-a^2+b^2)tan A=(a^2-b^2+c^2) tan B = (b^2-c^2+a^2)tanC

Grade:11

1 Answers

Arun
25763 Points
3 years ago
tan A = sinA/ cos A
tan A = a*k/[(b²+c²- a²)/2bc] = 2abc *k/(c² + b² -a²)
 
Similarly
Tan B = 2 abc *k/(a² +c² -b²)
 
Now
tan A/ tan B = (a² +c² -b²)/ (b² +c² -a²)
 
Similarly you can also prove other parts.
 
Regards
Arun (askIITians forum expert)

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