How do you establish the identity for the following trigonometric equation?

Question

${sec}^{2} u + {csc}^{2} u = sec u csc u$ $sec^2u+csc^2u=secucscu$

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Answer 1 · 2018-01-19T17:12:40

this identity is false, see below

${sec}^{2} x + {csc}^{2} x$ $sec^2x+csc^2x$

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

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

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

${sec}^{2} x {csc}^{2} x$ $sec^2xcsc^2x$