how to simplify (1/sin^2x) - (1/tan^2x)?

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how to simplify (1/sin^2x) - (1/tan^2x)?

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Answer 1 · 2018-04-06T19:51:47

$\frac{1}{{sin}^{2} x} - \frac{1}{{tan}^{2} x} = \frac{1}{{sin}^{2} x} - \frac{{cos}^{2} x}{{sin}^{2} x} = \frac{1 - {cos}^{2} x}{{sin}^{2} x}$ $1/sin^2x-1/tan^2x=1/sin^2x-cos^2x/sin^2x=(1-cos^2x)/sin^2x$

$= \frac{{sin}^{2} x}{{sin}^{2} x} = 1$ $=sin^2x/sin^2x=1$ .

Explanation:

Or, $\frac{1}{{sin}^{2} x} - \frac{1}{{tan}^{2} x} = {csc}^{2} x - {cot}^{2} x = 1$ $1/sin^2x-1/tan^2x=csc^2x-cot^2x=1$ .

Answer 2 · 2018-04-06T19:53:16

It happens to be exactly $1$ $1$ .

Explanation:

$(\frac{1}{{sin}^{2} x}) - (\frac{1}{{tan}^{2} x})$ $(1/sin^2x) - (1/tan^2x)$

$= (\frac{1}{{sin}^{2} x}) - ⎛ ⎝ \frac{1}{\frac{{sin}^{2} x}{{cos}^{2} x}} ⎞ ⎠$ $=(1/sin^2x) - (1/\frac{sin^2x}{\cos^2x})$

$= \frac{1}{{sin}^{2} x} - \frac{{cos}^{2} x}{{sin}^{2} x}$ $=1/sin^2x - \frac{\cos^2x}{sin^2x}$

$= \frac{1 - {cos}^{2} x}{{sin}^{2} x}$ $=\frac{1-cos^2x}{sin^2x}$

$= \frac{{sin}^{2} x}{{sin}^{2} x} = 1$ $=\frac{sin^2x}{sin^2x} = 1$

Note that this uses the Pythagorean trigonometric identity in the second-to-last step, which states that:

${sin}^{2} x + {cos}^{2} x = 1$ $sin^2x+cos^2x=1$

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how to simplify (1/sin^2x) - (1/tan^2x)?

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