Question

(16,17) тригонометрия

Answer 1

16)
$\frac{1 + {tg}^{4}a }{ {tg}^{2}a + {ctg}^{2}a } - {tg}^{2} a = \frac{ \frac{ {cos}^{4} a + {sin}^{4}a }{ {cos}^{4}a } }{ \frac{ {sin}^{2} a}{ {cos}^{2}a } + \frac{ {cos}^{2}a }{ {sin}^{2} a} } - {tg}^{2}a = \frac{\frac{ {cos}^{4} a + {sin}^{4}a }{ {cos}^{4}a }}{ \frac{ {sin}^{4}a + {cos}^{4} a }{ {sin}^{2}a \times {cos}^{2}a } } - {tg}^{2} a = \frac{ {sin}^{2} a \times {cos}^{2}a }{ {cos}^{4}a } - {tg}^{2}a = \frac{ {sin}^{2}a }{ {cos}^{2} a} - {tg}^{2} a = {tg}^{2} a - {tg}^{2} a = 0$
17)
${cos}^{2} (30 + x) + {cos}^{2} (30 - x) - {cos}^{2} x = {(cos30 cosx - sin30sinx)}^{2} + (cos30cosx + sin30sinx)^{2} - {cos}^{2} x = ( \frac{ \sqrt{3} }{2} cosx - \frac{1}{2} sinx)^{2} + {( \frac{ \sqrt{3} }{2} cosx + \frac{1}{2}sinx) }^{2} - {cos}^{2} x = \frac{3}{4} {cos}^{2} x - \frac{ \sqrt{3} }{2} cosxsinx + \frac{1}{4} {sin}^{2} x + \frac{3}{4} {cos}^{2} x + \frac{ \sqrt{3} }{2} cosxsinx + \frac{1}{4} {sin}^{2} x - {cos}^{2} x = \frac{6}{4} {cos}^{2} x + \frac{2}{4} {sin}^{2} x - {cos}^{2} x = \frac{2}{4} {cos}^{2} x + \frac{2}{4} {sin}^{2} x = \frac{2}{4} ( {cos}^{2} x + {sin}^{2} x) = \frac{2}{4} = \frac{1}{2} = 0.5$