Question

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Answer 1

$3) \ \dfrac{\text{ctg}^{2}\gamma - 1}{\text{ctg}^{2}\gamma + 1} - \cos^{2}\gamma = \dfrac{\text{ctg}^{2}\gamma - 1}{\dfrac{1}{\sin^{2}\gamma} } - \cos^{2}\gamma = (\text{ctg}^{2}\gamma - 1)\sin^{2}\gamma - \cos^{2}\gamma =\\= \text{ctg}^{2}\gamma \sin^{2}\gamma - \sin^{2}\gamma - \cos^{2}\gamma = \dfrac{\cos^{2}\gamma}{\sin^{2}\gamma}\sin^{2}\gamma -\sin^{2}\gamma - \cos^{2}\gamma =\\= \cos^{2}\gamma - \sin^{2}\gamma - \cos^{2}\gamma = -\sin^{2}\gamma$

$4) \ \dfrac{\text{tg}^{2}x - 1}{\text{tg}^{2}x + 1} - \sin^{2}x = \dfrac{\text{tg}^{2}x - 1}{\dfrac{1}{\cos^{2}x} } - \sin^{2}x = (\text{tg}^{2}x - 1)\cos^{2}x - \sin^{2}x =\\= \text{tg}^{2}x\cos^{2}x - \cos^{2}x - \sin^{2}x = \dfrac{\sin^{2}x}{\cos^{2}x} \cos^{2}x - \cos^{2}x - \sin^{2}x = \\= \sin^{2}x - \cos^{2}x - \sin^{2}x = -\cos^{2}x$