How do you prove ${tan}^{2} x {sin}^{2} x = {tan}^{2} x + {cos}^{2} x - 1$ $tan^2xsin^2x=tan^2x + cos^2x-1$ ?

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How do you prove ${tan}^{2} x {sin}^{2} x = {tan}^{2} x + {cos}^{2} x - 1$ $tan^2xsin^2x=tan^2x + cos^2x-1$ ?

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Answer 1 · 2015-05-21T15:56:51

${tan}^{2} x . {sin}^{2} x = {tan}^{2} x (1 - {cos}^{2} x) = {tan}^{2} x - {tan}^{2} x ({cos}^{2} x)$ $tan^2 x .sin^2 x = tan^2 x(1 - cos^2 x) = tan^2 x - tan^2 x(cos^2 x)$

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

$= {tan}^{2} x - 1 + {cos}^{2} x$ $= tan^2x - 1 + cos^2 x$

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How do you prove ${tan}^{2} x {sin}^{2} x = {tan}^{2} x + {cos}^{2} x - 1$ $tan^2xsin^2x=tan^2x + cos^2x-1$ ?

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How do you prove tan2xsin2x=tan2x+cos2x−1tan^2xsin^2x=tan^2x + cos^2x-1?

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How do you prove ${tan}^{2} x {sin}^{2} x = {tan}^{2} x + {cos}^{2} x - 1$ $tan^2xsin^2x=tan^2x + cos^2x-1$ ?