$\sqrt{3}\backslash sqrt3$ Tan $\theta \backslash theta$ = 1, then $\theta \backslash theta$ =

In a $\mathrm{△}ABC\backslash triangle\; ABC$ if $\mathrm{\angle }B\backslash angle\; B$ = 90° and Tan C = $\frac{5}{12}\backslash frac\{5\}\{12\}$ , then the length of the hypotenuse is

$Ta{n}^{2}60^{\circ }+4Co{s}^{2}45^{\circ }+3Se{c}^{2}30^{\circ }+5Co{s}^{2}90^{\circ }=Tan^2\; 60^\backslash circ\; +\; 4\; Cos^2\; 45^\backslash circ\; +\; 3\; Sec^2\; 30^\backslash circ\; +\; 5\; Cos^2\; 90^\backslash circ=$

$Se{c}^{2}27^{\circ }-Co{t}^{2}63^{\circ }=Sec^2\; 27^\backslash circ-Cot^263^\backslash circ=$

$\sqrt{\frac{1}{Se{c}^2\theta }+Si{n}^2\theta +\frac{1}{Ta{n}^2\theta }}=\backslash sqrt\{\backslash frac\{1\}\{Sec^2\; \backslash theta\}\; +\; Sin^2\; \backslash theta\; +\; \backslash frac\{1\}\{Tan^2\; \backslash theta\}\}\; =$

$Si{n}^2\theta (1+Co{t}^2\theta )=Sin^2\; \backslash theta\; (1\; +\; Cot^2\; \backslash theta)\; =$

If A = 45°, then $(1+TanA)(1+Ta{n}^{2}A)(1+Ta{n}^{3}A)(1\; +\; Tan\; A)\; (1\; +\; Tan^2\; A)\; (1\; +\; Tan^3\; A)$ =

$Si{n}^2\theta \mathrm{.}Co{t}^2\theta +Co{s}^2\theta \mathrm{.}Ta{n}^2\theta =Sin^2\; \backslash theta\; .\; Cot^2\; \backslash theta\; +\; Cos^2\; \backslash theta\; .\; Tan^2\; \backslash theta\; =$

$Tan45^{\circ }+2Ta{n}^{2}60^{\circ }=Tan\backslash ,45^\backslash circ+2\backslash ,Tan^260^\backslash circ=$

Sin 2A = 2 Sin A, then $\mathrm{\angle }A\backslash angle\; A$ =