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 $a = \cos2\alpha + \cos2\beta + 2\cos(\alpha + \beta )$ and $b = \sin2\alpha + \sin2\beta + 2\sin

Vedclass · Accepted Answer

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

Vedclass · Answer

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

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

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