Question

1/1+sin + 1/ 1-sin

Answer 1

\(\Huge\color{blue}{ \bf \frac{1}{1+sin(x)}+\frac{1}{1-sin(x)} }\) you can see that the denominators are conjugates of each other, right? So...

the least common denominator is
\[(1+\sin(x)~~)(~~1-\sin(x)~~)=1^2-\sin^2(x)=1-\sin^2(x)=Cos^2(x)\]

\(\Huge\color{blue}{ \bf \frac{1(1-sin(x)~~)}{Cos^2x}+\frac{1(1+sin(x)~~)}{Cos^2x} }\)

Answer 2

\(\Huge\color{blue}{ \bf \frac{1(1-sin(x)~~)}{Cos^2x}+\frac{1(1+sin(x)~~)}{Cos^2x} }\)


\(\Huge\color{blue}{ \bf \frac{1-sin(x)~~}{Cos^2x}+\frac{1+sin(x)~~}{Cos^2x} }\)


\(\Huge\color{blue}{ \bf \frac{1-sin(x)+1+sin(x)~~}{Cos^2x} }\)


\(\Huge\color{blue}{ \bf \frac{2+sin(x)~~}{Cos^2x} }\)

Answer 3

\(\Huge\color{blue}{ \bf \frac{2+sin(x)~~}{Cos^2x} }\)

\(\Huge\color{blue}{ \bf \frac{2~~}{Cos^2x}+\frac{sin(x)~~}{Cos^2x} }\)


\(\LARGE \color{blue}{ \bf 2sec^2x+tan(x)sec(x) }\)

Answer 4

that would be my final answer...