Question

secxtanx-cosxcotx=sinx
solve for x

Answer 1

pls help

Answer 2

after converting the whole eq. into sin ,cos you get
\[\sin ^{2}x - \cos ^{4}x=\sin ^{2}x.\cos ^{2}x\]
put \[\sin ^{2}x=1-\cos ^{2}x\] and solving, you get
\[1-\cos ^{2}x-\cos ^{4}x=\cos ^{2}x-\cos ^{4}x\]
cancelling \[-\cos ^{4}x\]
both sides
you get
\[\cos ^{2}x=\frac{ 1 }{ 2 }\]\[\cos ^{2}x=\cos ^{2}\frac{ \pi }{ 4 }\]\[x=n \pi \pm \frac{ \pi }{ 4 }\]
\[n \in integer\]