Question

prove that
$\frac{ \: \: \: {sin}^{2}a + 2cos \: a - 1 \: }{sin {}^{2}a \: + 3cos \: a - 3} = \frac{1}{1 - sec \: a}$
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Answer 1

Answer:

...$sin^2x =1 - cos^2x\\L.H.S: = \frac{(1-cos^2x)+2cosx -1}{(1-cos^2x)+3cosx -3} = \frac{-cos^2x+2cosx}{-2-cos^2x+3cosx} = \frac{cosx(cosx -2)}{(cosx -2)(cosx -1)} = \frac{cosx}{cosx-1} \\=\frac{1}{1-secx} = R.H.S$

Explanation: