Question

1 + sec^2(x) sin^2(x) = sec^2(x)

Answer 1

please show steps

Answer 2

\[\begin{align*}1+\sec^2x\sin^2x&=\sec^2x\left(\frac{1}{\sec^2x}+\sin^2x\right)\\\\
&=\sec^2x\left(\cos^2x+\sin^2x\right)\\
&=\sec^2x(1)\\
&=\sec^2x\end{align*}\]

Answer 3

how did you get 17sec^2(X)

Answer 4

1/sec^2(x)

Answer 5

Anything divided by itself is 1, right?
\[1=\frac{\sec^2x}{\sec^2x}=\sec^2x\left(\frac{1}{\sec^2x}\right)\]

Answer 6

ok i guess

Answer 7

im still confused

Answer 8

Is 2 times 1/2 equal to 1?

The answer's yes. What I did in the first step was work backwards - factorizing.
\[1=\frac{2}{2}=2\left(\frac{1}{2}\right)\]
Similarly,
\[1=\frac{\sec^2x}{\sec^2x}=\sec^2x\left(\frac{1}{\sec^2x}\right)\]
So, the \(1\) and \(\sec^2x\sin^2x\) can be written so that they share a common factor:
\[\begin{align*}1+\color{blue}{\sec^2x}\sin^2x&=\frac{\color{blue}{\sec^2x}}{\sec^2x}+\color{blue}{\sec^2x}\sin^2x\\
&=\color{blue}{\sec^2x}\left(\frac{1}{\sec^2x}+\sin^2x\right)\end{align*}\]

Answer 9

OMG THANK YOU SO MUCH

Answer 10

wait how did you get 1/sec^2(x) + sin^2(x)
where did that addition sign come from

Answer 11

The plus is on the left side of the equation:
\[1\color{red}+\sec^2x\sin^2x\]