Question

prove that
$(1 + \frac{1}{tan {}^{2}a } ) \: (1 + \frac{1}{cot {}^{2} a} ) = \frac{1}{sin {}^{2} a - sin {}^{4}a }$
using LHS =RHS
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plzz ans it fast
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Answer 1

Answer:

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Answer 2

Answer:

${\huge{\boxed{\red{\mathscr{SolutioN}}}}}$

$(1 + \frac{1}{tan {}^{2} a} )(1 + \frac{1}{cot {}^{2}a } )$

$= (1 + \frac{cos {}^{2} a}{sin {}^{2} a} )(1 + \frac{sin {}^{2}a }{1 + cos {}^{2}a } )$

$= ( \frac{sin {}^{2} a + cos {}^{2}a }{sin {}^{2} a} )( \frac{cos {}^{2}a + sin {}^{2} a}{cos {}^{2} a} )$

$= (\frac{1}{sin {}^{2}a } )( \frac{1}{cos {}^{2} a} )$

$= \frac{1}{sin {}^{2} a \: cos {}^{2} a}$

$= \frac{1}{sin {}^{2} a(1 - sin {}^{2} a)}$

$= \frac{1}{sin {}^{2} a - sin {}^{4} a}$