Question

Simplify (tanx + cosx + sinxtanx)/(secx + tanx) [answer: 1]

Could someone please explain to me how to get the answer?

Answer 1

convert all identies in sinx and cosx and then solve.i hope u will find ans urself

Answer 2

girlnotonfire do you happen to have your trig identities sheet?

Answer 3

Just the ones in my math book.

Answer 4

ok.... so you know what hmm say \(\bf cos^2(x)+sin^2(x)=?\) is right?

Answer 5

1

Answer 6

ok
so as WHassan suggested, let's expand the identities to just cosine and sine identities... so

Answer 7

one sec

Answer 8

u should not give direct ans

Answer 9

\(\bf \color{blue}{ tan(x)=\frac{sin(x)}{cos(x)}\qquad \qquad sec(x)=\frac{1}{cos(x)}\qquad cos^2(x)+sin^2(x)=1}\\ \quad \\ \quad \\

\cfrac{tan(x) + cos(x) + sin(x)tan(x)}{sec(x) + tan(x)}\implies \cfrac{{\color{blue}{ \frac{sin(x)}{cos(x)}}}+cos(x)+sin(x)\frac{sin(x)}{cos(x)}}{{\color{blue}{ \frac{1}{cos(x)}}}+\frac{sin(x)}{cos(x)}}\\ \quad \\

\implies \cfrac{
\frac{sin(x)}{cos(x)}+cos(x)+\frac{sin(x)^2}{cos(x)}
}{\frac{1}{cos(x)}+\frac{sin(x)}{cos(x)}}\implies
\cfrac{\frac{sin(x)+cos^2(x)+sin^2(x)}{cos(x)}}{\frac{1+sin(x)}{cos(x)}}
\)

Answer 10

\(\bf {\color{blue}{ tan(x)=\frac{sin(x)}{cos(x)}}\qquad \qquad sec(x)=\frac{1}{cos(x)}\qquad cos^2(x)+sin^2(x)=1}\\ \quad \\ \quad \\

\cfrac{
\frac{sin(x)}{cos(x)}+cos(x)+\frac{sin(x)^2}{cos(x)}
}{\frac{1}{cos(x)}+\frac{sin(x)}{cos(x)}}\implies
\cfrac{\frac{sin(x)+{\color{blue}{ cos^2(x)+sin^2(x)}}}{cos(x)}}{\frac{1+sin(x)}{cos(x)}}\)


so... notice that.... and replace by 1, then do the fractions, keeping in mind that \(\bf \cfrac{\quad \frac{a}{b}\quad }{\frac{c}{d}}\implies \cfrac{a}{b}\cdot \cfrac{d}{c}\)

Answer 11

Oh okay, that makes since. It simplifies to \[\cos^2x+\sin^2x\] so it equals 1.

Answer 12

sense*

Answer 13

\(\bf {\color{blue}{ tan(x)=\frac{sin(x)}{cos(x)}\qquad \qquad sec(x)=\frac{1}{cos(x)}\qquad cos^2(x)+sin^2(x)=1}}\\ \quad \\ \quad \\

\cfrac{
\frac{sin(x)}{cos(x)}+cos(x)+\frac{sin(x)^2}{cos(x)}
}{\frac{1}{cos(x)}+\frac{sin(x)}{cos(x)}}\implies
\cfrac{\frac{sin(x)+{\color{blue}{ cos^2(x)+sin^2(x)}}}{cos(x)}}{\frac{1+sin(x)}{cos(x)}}
\)

Answer 14

Thank you very much for your help!!

Answer 15

right... so .. \(\bf {\color{blue}{ tan(x)=\frac{sin(x)}{cos(x)}\qquad \qquad sec(x)=\frac{1}{cos(x)}\qquad cos^2(x)+sin^2(x)=1}}\\ \quad \\ \quad \\


\cfrac{\frac{sin(x)+{\color{blue}{ cos^2(x)+sin^2(x)}}}{cos(x)}}{\frac{1+sin(x)}{cos(x)}}\implies \cfrac{\frac{sin(x)+{\color{blue}{ 1}}}{cos(x)}}{\frac{1+sin(x)}{cos(x)}}\\ \quad \\
\implies

\cfrac{sin(x)+{\color{blue}{ cos^2(x)+sin^2(x)}}}{\cancel{cos(x)}}\cdot

\cfrac{\cancel{cos(x)}}{1+sin(x)}\)

Answer 16

woopps... anyhow I got the wrong ... fractoin... but \(\bf \cfrac{sin(x)+{\color{blue}{ 1}}}{\cancel{cos(x)}}\cdot

\cfrac{\cancel{cos(x)}}{1+sin(x)}\)