Question

Simplify and write the trigonometric expression in terms of sine and cosine. [(2+tan^2x)/(sec2x)]−1=(f(x))^2

Answer 1

1 + tan^2 = sec^2

also sec = 1/cos

\[(\frac{1 + \sec^2 x}{\sec 2x}) - 1 = \large (\frac{\frac{1+\cos^2 x}{\cos^2 x}}{\frac{1}{\cos 2x}}) - 1\]
cos 2x = cos^2 - sin^2
\[= \frac{(1+\cos^2 x)(\cos^2 x - \sin^2 x)}{\cos^2 x} - 1\]

Answer 2

thanks

Answer 3

They want the quantity to be the square of something. I believe that on can
proceed as follows:
\[
\frac{\tan ^2(x)+2}{\sec ^2(x)}-1=\frac{\tan ^2(x)+1+1}{\sec
^2(x)}-1=\\ \frac{\sec ^2(x)+1}{\sec ^2(x)}-1=\frac{1}{\sec
^2(x)}=\cos ^2(x)
\]

Answer 4

@lawls @dumbcow

Answer 5

ok thanks