Sin^2x+sin^2x cot^2x ?

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Sin^2x+sin^2x cot^2x ?

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Answer 1 · 2017-12-13T01:53:33

$1$ $1$

Explanation:

${sin}^{2} x + {sin}^{2} x {cot}^{2} x$ $sin^2x+sin^2x cot^2x$

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

$= {sin}^{2} x \cdot (\frac{{sin}^{2} x + {cos}^{2} x}{{sin}^{2} x})$ $= sin^2x*((sin^2x+cos^2x)/sin^2x)$

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

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

$= 1$ $= 1$

Answer 2 · 2017-12-13T01:55:47

See the answer below....

Explanation:

${sin}^{2} x + {sin}^{2} x \cdot {cot}^{2} x$ $sin^2x+sin^2x cdot cot^2x$
$= {sin}^{2} x + {sin}^{2} x \cdot \frac{{cos}^{2} x}{{sin}^{2} x}$ $=sin^2x+sin^2x cdot cos^2x/sin^2x$
$=sin^2x+cancel(sin^2x) cdot cos^2x/cancel(sin^2x)$
$=sin^2x+cos^2x$
$=1$ [As identity]

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Sin^2x+sin^2x cot^2x ?

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