Question

Find in the missing step:
sec^4 x -2 sec^2 x+1=(sec^2x-1)^2
=?????????
=tan^4 x

Answer 1

(tan^2x)^2

Answer 2

[ sec^2x-1= tan^2x ]

Answer 3

how is that?

Answer 4

is it because tan^2x=sec^2x-1?

Answer 5

Yes

Answer 6

its an identity =)

Answer 7

okay thanks

Answer 8

So, the best way to do this is to go back to basics of a right triangle!!



H = hypotenuse, B = base, P = perpendicular

\[\sec(x) = \frac{H}{B}\]

\[\sec^2(x) = \frac{H^2}{B^2}\]

\[Sec^2(x) - 1 = \frac{H^2}{B^2} - 1 = \frac{H^2 - B^2}{B^2}\]

Now, according to the Pythagoras theorem,
\[H^2 = P^2 + B^2\]

Which means
\[H^2 - B^2 = P^2\]

So,
\[Sec^2(x) - 1 = \frac{H^2 - B^2}{B^2} = \frac{P^2}{B^2}\]

\[\frac{P}{B} = tan(x)\]

Thus, you get your answer :)