Question

If cot B=12/5,prove that tan2B-sin2B=sin4B sec2B

Answer 1

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)

Answer 2

Answer:

Step-by-step explanation: