Question

Prove L.H.S = R.H.S
I will Mark Him BRAINLIEST Who Prove This Sums.
class 10 ML. Aggarwal​

Answer 1

$\huge{\bigstar}}$$\huge{\boxed{\red{\mathfrak{Answer}}}}$$\huge{\bigstar}}$

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$\mathtt{L.H.S}$

$= \frac{sin \: x}{ \frac{cos \: x}{sin \: x} + \frac{1}{sin\: x} } \\ \\ = \frac{sin \: x}{ \frac{1 + cos \: x}{sin \: x} } \\ \\ = \frac{ {sin}^{2} \: x }{1 + cos \: x} \\ \\ = \frac{ 1 - {cos}^{2} x}{1 + cos \: x} \\ \\ = \frac{(1 - cos \: x)(1 + cos \: x)}{1 + cos \: x} \\ = 1 - cos \: x$

$\mathtt{R.H.S}$

$= \frac{sin \: x}{ \frac{1 - cos \: x}{sin \: x} } + 2 \\ \\ = \frac{ {sin}^{2} \: x}{1 - cos \: x} + 2 \\ \\ similarly \\ = 1 + cos \: x + 2 \\ = 3 + cos \: x$

Hence that's why

L.H.S ≠ R.H.S