If cos = x^2 - y^2x^2 + y^2 then the value ofcot is equal to [ if 0^ 90^ ]

cos = x^2 - y^2x^2 + y^2 AC2 = (x2 + y2)2 - (x2 - y2)2 = x4 + y4 + 2x2y2 - x4 - y4 + 2x2y2 = 4x2y2 ⇒ AC = 2xy ⇒ cotθ = x^2 - y^22xy

This question belongs to Arithmetic Ability Trigonometry

Trigonometry

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 & Solution

Correct Answer: Option D

$${\text{cos}}\theta = \frac{{{x^2} - {y^2}}}{{{x^2} + {y^2}}}$$
Trigonometry mcq solution image

AC² = (x² + y²)² - (x² - y²)²
= x⁴ + y⁴ + 2x²y² - x⁴ - y⁴ + 2x²y²
= 4x²y²
⇒ AC = 2xy
⇒ cotθ = $$\frac{{{x^2} - {y^2}}}{{2xy}}$$

Question posted by Examveda

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