Question

please give me answer​

Answer 1

Step-by-step explanation:

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)

thank you hope it helps

Answer 2

Answer:

otB=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)

thank you hope it helps

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