How do you simplify ${(\frac{sin x}{cos x})}^{2} - \frac{1}{{cos}^{2} x}$ $(sinx/cosx)^2-1/cos^2x$ ?

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Answer 1 · 2016-09-06T02:52:32

We have that

${(\frac{sin x}{cos x})}^{2} - \frac{1}{{cos}^{2} x}$ $(sinx/cosx)^2-1/cos^2x$

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

$- \frac{1}{{cos}^{2} x} \cdot (1 - {sin}^{2} x)$ $-1/(cos^2x)*(1-sin^2x)$

$- \frac{1}{{cos}^{2} x} \cdot {cos}^{2} x$ $-1/(cos^2x)*cos^2x$

$- 1$ $-1$

How do you simplify (sinxcosx)2−1cos2x(sinx/cosx)^2-1/cos^2x?