Question

plse solve this

Answer 1

\[\tan ^{2}A=\cos ^{2}B - \sin ^{2} B\]

Answer 2

like that?

Answer 3

yes

Answer 4

\[\frac{ \sin ^{2}A }{\cos ^{2}A }=\frac{ \cos ^{2}B-\sin ^{2}B }{1 }\]

Answer 5

apply componendo and dividendo

Answer 6

\[\frac{ \cos ^{2}A-\sin ^{2}A }{\cos ^{2}A+\sin ^{2}A }=\frac{ 1-\left( \cos ^{2}B-\sin ^{2}B \right) }{ 1+\left( \cos ^{2}B+\sin ^{2}B \right) }\]

Answer 7

\[\frac{ \cos ^{2}A-\sin ^{2}A }{1 }=\frac{ 1-\cos ^{2}B+\sin ^{2}B }{1-\sin ^{2} B+\cos ^{2}B }\]

Answer 8

\[\cos ^{2}A-\sin ^{2}A=\frac{ \sin ^{2}B+\sin ^{2}B }{\cos ^{2}B+\cos ^{2}B }\]
Now you can solve.