secx^2-1/sinx=sinx/1-sin^2x

Question

mathmale · Answer

aatka:  Hope you won't take my input too negatively:  I strongly suggest that you include with all these equations and symbols the directions we're supposed to follow in "solving" this problem.  Secondly, I'd suggest that when dividing by expressions such as 1-(sin x)^2, you enclose those expressions in parentheses so that there's no doubt as to what your divisor includes and does not include:  [1-(sin x)^2].

Please type in the instructions for this problem; perhaps then I could give you some help.

anonymous · Answer

Is this your problem
$$
\frac{\sec ^2(x)-1}{\sin
   (x)}=\frac{\sin (x)}{1-\sin
   ^2(x)}
$$
To show that they are equal

anonymous · Answer

Notice that
$$
\frac{\sec ^2(x)-1}{\sin
   (x)}=\tan (x) \sec (x)

$$
Can you show that the second side is also equal to
$$
\tan (x) \sec (x)

$$

anonymous · Answer

You can show easily that
$$
\frac{\sin (x)}{1-\sin
   ^2(x)}=\tan (x) \sec (x)
$$