How do you verify $\frac{- 1}{tan x - sec x} = \frac{1 + sin x}{cos x}$ $(-1) / (tanx - secx) = (1 + sinx) / (cosx)$ ?

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How do you verify $\frac{- 1}{tan x - sec x} = \frac{1 + sin x}{cos x}$ $(-1) / (tanx - secx) = (1 + sinx) / (cosx)$ ?

1 Answer

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Answer 1 · 2015-11-04T05:04:21

Verify trig expression.

Explanation:

Develop the left side.

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

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

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How do you verify $\frac{- 1}{tan x - sec x} = \frac{1 + sin x}{cos x}$ $(-1) / (tanx - secx) = (1 + sinx) / (cosx)$ ?

1 Answer

Explanation:

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How do you verify (-1) / (tanx - secx) = (1 + sinx) / (cosx)−1tanx−secx=1+sinxcosx (-1) / (tanx - secx) = (1 + sinx) / (cosx)?

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Explanation:

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How do you verify $\frac{- 1}{tan x - sec x} = \frac{1 + sin x}{cos x}$ $(-1) / (tanx - secx) = (1 + sinx) / (cosx)$ ?