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