Question

sinx*tanx=1-cos^2x/cosx

Answer 1

\(\LARGE\color{blue}{ \bf sin~x~tan~x~=~\frac{1-cos^{2}x}{cos~x} }\)
like this ?

Answer 2

\[\sin(x)\tan(x)=\frac{ \sin(x) }{ \cos(x) }\]
\[\[\sin(x)\tan(x)=\tan(x)\]
\[\sin(x)=1\]
\[x=\sin ^{-1}(1)=90\]

Answer 3

No no no !!!!

\(\LARGE\color{blue}{ \bf sin~x~tan~x~=~\frac{1-cos^{2}x}{cos~x} }\)

\(\LARGE\color{blue}{ \bf sin~x~tan~x~=~\frac{sin^{2}x}{cos~x} }\)

\(\LARGE\color{blue}{ \bf sin~x~tan~x~=sin~x~~\frac{sinx}{cos~x} }\)

\(\LARGE\color{blue}{ \bf sin~x~tan~x~=sin~x~tan~x~ }\)

Answer 4

Ohh, that's true I forgot it is equal to sin^2(x) xD

Answer 5

Yeah :)