Question

tan37 1/2= √6+√3-√2-2​

Answer 1

Answer:

i am sorry idk but pls answer this

Step-by-step explanation:

5∠A = 3∠B = ∠C, ∠A and ∠B are complementary angles, then the measure of ∠A, ∠B and

∠C are respectively

Answer 2

we know that

$\frac{1 - \cos(75) }{1 + \cos(75) } = \tan {}^{2} ( \frac{75}{2} ) .....(i)$

Now,

$\cos(75) = \cos(45 + 30) = \frac{ \sqrt{3} - 1}{2 \sqrt{2} }$

From (i),

$\tan {}^{2} ( \frac{75}{2} ) = \frac{1 - \frac{ \sqrt{3} - 1 }{2 \sqrt{2} } }{1 + \frac{ \sqrt{3} - 1 }{2 \sqrt{2} } } = \frac{2 \sqrt{2} - \sqrt{3} + 1}{2 \sqrt{2} + \sqrt{3} - 1 } =\frac{2 \sqrt{2} - \sqrt{3} + 1}{2 \sqrt{2} + \sqrt{3} - 1 } \times \frac{2 \sqrt{2} - \sqrt{3} + 1}{2 \sqrt{2} - \sqrt{3} + 1 } = \frac{(2 \sqrt{2} - \sqrt{3} + 1) {}^{2} }{(2 \sqrt{2}) {}^{2} - ( \sqrt{3} - 1) {}^{2} } = \frac{(2 \sqrt{2} - \sqrt{3} + 1) {}^{2} }{8 - 3 - 1 + 2 \sqrt{3} } = \frac{(2 \sqrt{2} - \sqrt{3} + 1) {}^{2} }{4 + 2 \sqrt{3} } = \frac{(2 \sqrt{2} - \sqrt{3} + 1) {}^{2} }{( \sqrt{3} + 1) {}^{2} }$

$\tan( \frac{75}{2} ) = \frac{(2 \sqrt{2} - \sqrt{3} + 1) }{ \sqrt{3} + 1 } = \frac{(2 \sqrt{2} - \sqrt{3} + 1) }{ \sqrt{3} + 1 } \times \frac{ \sqrt{3} - 1}{ \sqrt{3} - 1} = \frac{2 \sqrt{6} - 3 + \sqrt{3} - 2 \sqrt{2} + \sqrt{3} - 1 }{3 - 1} = \sqrt{6 } + \sqrt{3} - \sqrt{2} - 2$