# Math Help - proving trigonometric identity

1. ## proving trigonometric identity

Write the following expression in either sine or cosine

$1) sin^2x+ cos^2x * cscx$

Attempt:
$sin^2 + cos^2 * \frac{1}{sinx}$
$sin^2 + \frac{cos^2}{sinx}$
$\frac{sin^2x}{1} + \frac{cos^2}{sinx}$
$\frac{sin^3x+cos^2}{sinx}$
$sin^3 +cos^2 * \frac{1}{sinx}$
$1*1=1$

$cosx +sin^2x* secx$

Are my above steps correct please write in step if they are incorrect.
Thank you very much!

2. Originally Posted by mj.alawami
Write the following expression in either sine or cosine

$1) sin^2x+ cos^2x * cscx$

Attempt:
$sin^2 + cos^2 * \frac{1}{sinx}$
$sin^2 + \frac{cos^2}{sinx}$
$\frac{sin^2x}{1} + \frac{cos^2}{sinx}$
$\frac{sin^3x+cos^2}{sinx}$
$sin^3 +cos^2 * \frac{1}{sinx}$
$1*1=1$

$\frac{sin^3x+cos^2}{sinx}$

$
=\frac{sin^3x-sin^2x+1}{sinx}
$

3. Originally Posted by mj.alawami
Write the following expression in either sine or cosine

$1) sin^2x+ cos^2x * cscx$

Attempt:
$sin^2 + cos^2 * \frac{1}{sinx}$
$sin^2 + \frac{cos^2}{sinx}$
$\frac{sin^2x}{1} + \frac{cos^2}{sinx}$
$\frac{sin^3x+cos^2}{sinx}$
_______________________________________
$(sin^3 +cos^2 )* \frac{1}{sinx}$
_______________________________________
___________________________
$1*1=1$
__________________________

$cosx +sin^2x* secx$

Are my above steps correct please write in step if they are incorrect.
Thank you very much!
-First of all watch the step in Red (correct)

-Now watch the step in Green (incorrect)

If your question is (doubtful since not a very good use of bracket)

$cosx +sin^2x* secx$

$= \frac{sin^2(x) }{cos(x)} +cos(x)$

$
= \frac{sin^2(x) +cos^2(x)}{cos(x)}

$

Since
$
sin^2(x) +cos^2(x) =1$

$
= \frac{1}{cos(x)}
$