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 $a$ and $b$.$a = \sin 2\alpha + \sin 2\beta + 2\sin(\alpha + \beta) = 2\sin(\alpha + \beta)\cos(\

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

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

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

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