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यदि cos 43° = $\frac{x}{\sqrt{x^{2} + y^{2}}}$ है, तो tan 47° का मान है –

1
$\frac{y}{\sqrt{x^{2} + y^{2}}}$
2
$\frac{y}{x}$
3
$\frac{x}{y}$
4
$\frac{x}{\sqrt{x^{2} - y^{2}}}$

Answer:- 3

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