1. proving trigonometric identity

Write the following expression in either sine or cosine

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

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

$\displaystyle cosx +sin^2x* secx$

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

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

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

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

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

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

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

Attempt:
$\displaystyle sin^2 + cos^2 * \frac{1}{sinx}$
$\displaystyle sin^2 + \frac{cos^2}{sinx}$
$\displaystyle \frac{sin^2x}{1} + \frac{cos^2}{sinx}$
$\displaystyle \frac{sin^3x+cos^2}{sinx}$
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$\displaystyle (sin^3 +cos^2 )* \frac{1}{sinx}$
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$\displaystyle 1*1=1$
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$\displaystyle cosx +sin^2x* secx$

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)

$\displaystyle cosx +sin^2x* secx$

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

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

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

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