Question

Solve above questions​

Answer 1

Answer:

hey mate here is urs answer

Answer 2

$\bold{\large{\underline{\underline{\orange{\mathfrak{Question\:23}}}}}} \\ \\ \tt Quadratic\:polynomial\:with\:zeroes\:5-3 \sqrt{2} \: and \: 5 + 3 \sqrt{2} \\ \\ \tt Sum \:of\: zeroes = 5 - 3 \sqrt{2} + 5 + 3 \sqrt{2} \\ \\ \tt \therefore \boxed{\bold{\underline{\green{\tt Sum\:of\:zeroes = 10 }}}} \\ \\ \tt Product\:of\:zeroes = ( 5 + 3\sqrt{2} ) ( 5 - 3\sqrt{2}) \\ \\ \tt \therefore Product\:of\:zeroes = {5}^{2} - {3\sqrt{2}}^{2} \\ \\ \tt Product\:of\:zeroes = 25 - 18 \\ \\ \tt \therefore \boxed{\bold{\underline{\green{\tt product \:of \:zeroes = 7 }}}} \\ \\ \tt Now\:the\: quadratic\:equation\:is\:written\:by \\ \\ \tt \therefore {x}^{2} - (sum\:of\:roots)x + product\:of\: roots = 0 \\ \\ \tt Therefore\:required\: quadratic\:equation\:is \\ \\ \tt \therefore \boxed{\bold{\underline{\red{\tt {x}^{2} - 10x + 7 = 0 }}}} \\ \\ \tt \bold{\large{\underline{\underline{\orange{\mathfrak{Question\:25}}}}}} \\ \\ \tt tanA = \frac{3}{4} \\ \\ \tt \therefore sinA = \frac{3}{5} \:\:\:\:\:\:\:\: cosA = \frac{4}{5} \\ \\ \tt \frac{1}{sinA} + \frac{1}{cosA} \\ \\ \tt \therefore \frac{5}{3} + \frac{5}{4} \\ \\ \tt \therefore \frac{20 + 15}{12} \\ \\ \tt \therefore \boxed{\bold{\underline{\red{\tt \frac{1}{sinA} + \frac{1}{cosA} = \frac{35}{12} }}}} \\ \\ \tt \sqrt{3}sin\theta - cos\theta = 0 \\ \\ \tt \sqrt{3}sin\theta = cos\theta \\ \\ \tt \therefore tan\theta = \frac{1}{\sqrt{3}} \\ \\ \tt \therefore \boxed{\bold{\underline{\red{\tt \theta = \frac{\pi}{6} }}}}$