How do you prove that ${tan}^{2} x = {csc}^{2} x \cdot {tan}^{2} x - 1$ $tan^2x = csc^2x * tan^2x - 1$ ?

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Answer 1 · 2016-10-16T18:19:34

Apply the identities $tan θ = \frac{sin θ}{cos θ}$ $tantheta = sintheta/costheta$ and $csc θ = \frac{1}{sin θ}$ $csctheta = 1/sintheta$ .

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

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

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

Apply the identity $1 - {cos}^{2} x = {sin}^{2} x$ $1 - cos^2x = sin^2x$ ( ${sin}^{2} x + {cos}^{2} x = 1$ $sin^2x+ cos^2x = 1$ ).

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

$L H S = R H S$ $LHS = RHS$

Hopefully this helps!

How do you prove that tan^2x = csc^2x * tan^2x - 1tan2x=csc2x⋅tan2x−1tan^2x = csc^2x * tan^2x - 1?