If $${ ext{cos}} heta= \frac{{{x^2} - {y^2}}}{{{x^2} + {y^2}}}$$then the value of$${ ext{cot}} heta $$is equal to $$\left[ {{ ext{if }}{0^ \circ } \leqslant heta\leqslant {{90}^ \circ }} ight]$$

Question

If $${	ext{cos}}	heta= \frac{{{x^2} - {y^2}}}{{{x^2} + {y^2}}}$$then the value of$${	ext{cot}}	heta $$is equal to $$\left[ {{	ext{if }}{0^ \circ } \leqslant 	heta\leqslant {{90}^ \circ }} ight]$$

TuteeHUB · Accepted Answer

NA

TuteeHUB · Answer

$\frac{{2xy}}{{{x^2} - {y^2}}}$$

TuteeHUB · Answer

$\frac{{2xy}}{{{x^2} + {y^2}}}$$

TuteeHUB · Answer

$\frac{{{x^2} + {y^2}}}{{2xy}}$$

TuteeHUB · Answer

$\frac{{{x^2} - {y^2}}}{{2xy}}$$

1.	If $${\text{cos}}\theta= \frac{{{x^2} - {y^2}}}{{{x^2} + {y^2}}}$$then the value of$${\text{cot}}\theta $$is equal to $$\left[ {{\text{if }}{0^ \circ } \leqslant \theta\leqslant {{90}^ \circ }} \right]$$
A.	$\frac{{2xy}}{{{x^2} - {y^2}}}$$
B.	$\frac{{2xy}}{{{x^2} + {y^2}}}$$
C.	$\frac{{{x^2} + {y^2}}}{{2xy}}$$
D.	$\frac{{{x^2} - {y^2}}}{{2xy}}$$
Answer» E.

If $${\text{cos}}\theta= \frac{{{x^2} - {y^2}}}{{{x^2} + {y^2}}}$$then the value of$${\text{cot}}\theta $$is equal to $$\left[ {{\text{if }}{0^ \circ } \leqslant \theta\leqslant {{90}^ \circ }} \right]$$

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