Question

plzzz answer this fast its so urgent.. ans fast ill mark as brainliest

Answer 1

As per the problem,
To prove -
$\frac{sin \: x}{1 - cos \: x} + \frac{tan \: x}{1 + cos \: x} = sec \: x.cosec \: x + cot \: x$
Taking LHS,

After simplifying,
$\frac{sin \: x \times (1 + cos \: x) + tan \: x \times (1 - cos \: x)}{1 - {cos \: }^{2} x}$
$\frac{sin \: x+ sin \: x.cos \: x + tan \: x - sin \: x}{{sin \: }^{2} x}$
As we know,
$tan \: x = \frac{sin \: x}{cos \: x} \\ {sin}^{2}x + {cos}^{2}x \: = 1$
$\frac{sin \: x.cos \: x + tan \: x}{{sin \: }^{2} x} \\ \frac{sin \: x.cos \: x}{ {sin}^{2} x} + \frac{tan \: x}{ {sin}^{2} x}$

$\frac{sin \: x.cos \: x}{ {sin}^{2} x} = cot \: x \: and \: \frac{tan \: x}{ {sin}^{2} x} = \frac{sin \: x}{cos \: x. {sin}^{2}x } = cosec \: x.sec \: x$

As we get,

$cosec \: x.sec \: x \: + cot \: x$

Hope it helped you!
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