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If $cosec θ=\frac{(x^2+y^2)}{(x^2-y^2)}$, then what will be the value of tan θ ? |
$\frac{(x^2-y^2)}{(x^2+y^2)}$ $\frac{2xy}{(x^2-y^2)}$ $\frac{(x^2-y^2)}{2xy}$ $\frac{(x^2+y^2)}{2xy}$ |
$\frac{(x^2-y^2)}{2xy}$ |
cosec θ = \(\frac{x² + y² }{x² - y²}\) { we know , cosec θ = \(\frac{H}{P}\) } By using pythagoras theorem, P² + B² = H² (x² - y²)² + B² = (x² + y²)² On solving it, B = 2xy Now, tan θ = \(\frac{P}{B}\) = \(\frac{x² - y²}{2xy}\) |
