Sin $\theta \backslash theta$ = $\frac{20}{29}\backslash frac\{20\}\{29\}$ then $Si{n}^2\theta +co{s}^2\theta Sin^2\; \backslash theta\; +\; cos^2\; \backslash theta$ =

If $\theta \backslash theta$ = 45° then the value of $\frac{1-Cos2\theta }{Sin2\theta }\backslash frac\{1-Cos\; 2\; \backslash theta\}\{Sin\; 2\; \backslash theta\}$ is

In $\mathrm{△}ABC\backslash triangle\; ABC$ if Sin A = $\frac{9}{15}\backslash frac\{9\}\{15\}$ then $Cose{c}^{2}A-co{t}^{2}A=Cosec^2A-cot^2A=$

$(1+Ta{n}^2{45}^{\circ }{)}^2=(1\; +\; Tan^2\; 45^\backslash circ)^2=$

$Si{n}^2{105}^{\circ }+co{s}^2{105}^{\circ }Sin^2\; 105^\backslash circ+cos^2105^\backslash circ$

Tan (B + 15°) = $\frac{1}{\sqrt{3}}\backslash frac\{1\}\{\backslash sqrt3\}$ then B =

Maximum value of Sin $\theta \backslash theta$  =

If Sec 2A = Cosec (A – 27°), then the value of $\mathrm{\angle }A\backslash angle\; A$ is

Sin (A – B) = $\frac{1}{2}\backslash frac\{1\}\{2\}$ ; Cos (A + B) = $\frac{1}{2}\backslash frac\{1\}\{2\}$ then $\mathrm{\angle }A\backslash angle\; A$ is

$Ta{n}^2{60}^{\circ }+2Ta{n}^2{45}^{\circ }=xTan{45}^{\circ }Tan^2\; 60^\backslash circ+2Tan^245^\backslash circ=x\backslash ,Tan45^\backslash circ$Then $x=x=$