can somebody please solve this for me?if A+B+C=180* tan^2A+tan^2B=tanAtanBtanC(cosec2A-cosec2B)?

Question

A+B+C=180*.
Please help me prove that
tan^2A+tan^2B
=tanAtanBtanC
(cosec2A-cosec2B)

Answer 1

Take a triangle A=45, B=60, C=75
tan^2(A) +tan^2(B) = 1 + 3 = 4

tan(A)*tan(B)*tan(C) *( csc(2A)-csc(2B) )
3.732050808*(1- 1.154700538)
3.732050808*(-0.154700538)
-0.577350268

leftside of the equation = 4 does not equal the rightside of the equation = -0.577..

So this equation is incorrect

Answer 2

Since A+B+C=180, C=180-(A+B).
Use tan(C)=-tan(A+B) and expand all the terms. Simplify to the required form.
Working out a problem is more important in mathematics than to know the steps involved to solve.

Answer 3

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Answer 4

I think the above is incorrect as LHS is symetric in A B and right is not

Answer 5

Try to use this:

tan^2A=(tg(2A)-2tgA)/tg(2A)

Actually mathdumb this is correct
You get it from the formula tg(a+b)=(tga+tgb)/(1-tgatgb)

You do the calculus mathdumb

Answer 6

Actually a small child can also give answer of this. So plz. ask him and dont waste our and yours time.

Answer 7

i did AP calculus my junior year of high school.....blah...im done with that