If cot B =12/5,show that tan ^2 B -sin^2 B=sin^2 B tan ^2 B
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Given that, CotB=12/5
We take a traigle
CotB=base/perpendicular
=12/5
By using Pythagoras theorem;
12²+5²=144+25=169
hypotenuse=√169=13
LHS:tan²B-sin²B
=(5/12)²-(5/13)²
= 25/144 - 25/169
= {25(169-144)}/144×169
=25×25/144×169
=625/24,336
Similarly;RHS:
=sin²Btan²B
=(5/13)²(5/12)²
=25/169×25/144
=25×25/169×144
=625/24,336
LHL=RHL [Proved]
I hope it is helpful.....
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