Question

Cot B=12÷5,show that tan²B-sin²B =sin²B - tan²B

Answer 1

Given cot b = 12/5 then tan b = 5/12 and sin b = 5/13, sec b = 13/12.

LHS = tan^2 b - sin^2 b

       = (5/12)^2 - (5/13)^2

      = 5^2(1/12^2 - 1/13^2)

      = 25(1/144 - 1/169)

      = 25(169 - 144/144 * 169)

      = 25(25/144 * 169)

     = 625/24336


RHS = sin^2b - tan^2b

         = (5/13)^2 - (5/12)^2

         = 5^2(1/13^2 - 1/12^2)

         = 25(169 - 144/169 * 144)

         = 625/24336.


LHS = RHS.


Hope this helps!