Question

CotB=12/5 . Prove that tan2 B - sin4 B=sin4 B*sec2 B

Answer 1

Answer:

cotB=12/5=base/perpendicular

By Pythagorus's theorem, p²+b²=h²

Here, p=5, b=12

∴, h=√(5²+12²)=√(25+144)=√169=13

∴tanB=p/b=5/12, sinB=p/h=5/13, secB=13/12

∴, tan²B-sin²B

=(5/12)²-(5/13)²

=25/144-25/169

=25{(169-144)/24336}

=625/24336

sin⁴Bsec²B

=(5/13)⁴×(13/12)²

=5⁴/13²×1/12²

=625/169×144

=625/24336

∴, LHS=RHS (Proved)

i hope its helpful