If-tan-pi-24-a-b-c-d-where-a-b-c-d-are-postive-numbers-Find-the-value-of-a-b-c-d-2-

Question Number 194937 by dimentri last updated on 20/Jul/23

I f \tan (\frac{π}{24}) = (\sqrt{a} - \sqrt{b}) (\sqrt{c} - \sqrt{d})

w h e r e a, b, c, d a r e p o s t i v e n u m b e r s .

F i n d t h e v a l u e o f (a + b + c + d + 2)

Answered by BaliramKumar last updated on 20/Jul/23

\tan (\frac{π}{24}) = \tan 7 . 5 °

\tan 15 ° = 2 - \sqrt{3}

\frac{2 \tan 7 . 5 °}{1 - \tan^{2} 7.5 °} = 2 - \sqrt{3} \Rightarrow \frac{2 x}{1 - x^{2}} = 2 - \sqrt{3}

x = \tan 7 . 5 = 2 \sqrt{2 + \sqrt{3}} - 2 - \sqrt{3}

\tan 7 . 5 = \sqrt{2} \sqrt{4 + 2 \sqrt{3}} - 2 - \sqrt{3}

\tan 7 . 5 = \sqrt{2} (\sqrt{3} + 1) - 2 - \sqrt{3}

\tan 7 . 5 = \sqrt{6} + \sqrt{2} - 2 - \sqrt{3}

\tan 7 . 5 = \sqrt{6} - \sqrt{3} - 2 + \sqrt{2}

\tan 7 . 5 = \sqrt{3} (\sqrt{2} - 1) - \sqrt{2} (\sqrt{2} - 1)

(\sqrt{a} - \sqrt{b}) (\sqrt{c} - \sqrt{d}) = (\sqrt{3} - \sqrt{2}) (\sqrt{2} - \sqrt{1})

a = 3, b = 2, c = 2, d = 1

a + b + c + d + 2 = \begin{matrix} 10 \end{matrix}