If the sides of a right-angled triangle are { | vedclass

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(D) Let the two legs of the right-angled triangle be $a = cos2\alpha + cos2\beta + 2cos(\alpha + \beta )$ and $b = sin2\alpha + sin2\beta + 2sin(\alph

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Both $(a)$ and $(c)$

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(D) Let the two legs of the right-angled triangle be $a = cos2\alpha + cos2\beta + 2cos(\alpha + \beta )$ and $b = sin2\alpha + sin2\beta + 2sin(\alpha + \beta )$.Using the sum-to-product formulas:$a = 2cos(\alpha + \beta)cos(\alpha - \beta) + 2cos(\alpha + \beta) = 2cos(\alpha + \beta)[cos(\alpha - \beta) + 1] = 4cos(\alpha + \beta)cos^2\left(\frac{\alpha - \beta}{2}ight)$.$b = 2sin(\alpha + \beta)cos(\alpha - \beta) + 2sin(\alpha + \beta) = 2sin(\alpha + \beta)[cos(\alpha - \beta) + 1] = 4sin(\alpha + \beta)cos^2\left(\frac{\alpha - \beta}{2}ight)$.The hypotenuse $h = \sqrt{a^2 + b^2}$.$h = \sqrt{[4cos^2\left(\frac{\alpha - \beta}{2}ight)]^2 [cos^2(\alpha + \beta) + sin^2(\alpha + \beta)]}$.Since $cos^2(\alpha + \beta) + sin^2(\alpha + \beta) = 1$,we have $h = 4cos^2\left(\frac{\alpha - \beta}{2}ight)$.Also,using the identity $2cos^2	heta = 1 + cos2	heta$,we have $4cos^2\left(\frac{\alpha - \beta}{2}ight) = 2[1 + cos(\alpha - \beta)]$.Thus,both $(a)$ and $(c)$ are correct.

If the sides of a right-angled triangle are ${cos2\alpha + cos2\beta + 2cos(\alpha + \beta )}$ and ${sin2\alpha + sin2\beta + 2sin(\alpha + \beta )}$,then the length of the hypotenuse is:

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