$\sqrt{1+SinA}\mathrm{.}\sqrt{1-SinA}=\backslash sqrt\{1+Sin\; A\}.\backslash sqrt\{1-Sin\; A\}\; =$

Sin $\theta \backslash theta$ = $\frac{a}{b}\backslash frac\{a\}\{b\}$ ; Cos $\theta \backslash theta$ = $\frac{c}{d}\backslash frac\{c\}\{d\}$ then Cot $\theta \backslash theta$ =

Sin $\theta \backslash theta$ . Cot $\theta \backslash theta$ . Sec $\theta \backslash theta$ =

$\frac{\sqrt{1+Ta{n}^2\theta }}{\sqrt{1+Co{t}^2\theta }}=\backslash frac\{\backslash sqrt\{1+Tan^2\; \backslash theta\}\}\{\backslash sqrt\{1+Cot^2\; \backslash theta\}\}\; =$

$Si{n}^3\theta Cos\theta +Co{s}^3\theta Sin\theta =Sin^3\; \backslash theta\; \backslash ,\; Cos\; \backslash theta\; +\; Cos^3\; \backslash theta\; \backslash ,\; Sin\; \backslash theta\; =$

Tan $\theta \backslash theta$ + Cot $\theta \backslash theta$ = 2, then $Ta{n}^2\theta +Co{t}^2\theta Tan^2\; \backslash theta\; +\; Cot^2\; \backslash theta$ =

$\frac{1+Sin\theta }{\sqrt{1-Si{n}^2\theta }}=\backslash frac\{1+Sin\; \backslash theta\}\{\backslash sqrt\{1-Sin^2\; \backslash theta\}\}\; =$

$\frac{Sin\alpha }{Cos\alpha }\mathrm{.}Cot\alpha =\backslash frac\{Sin\backslash ,\; \backslash alpha\}\{Cos\backslash ,\; \backslash alpha\}\; .\; Cot\backslash ,\; \backslash alpha\; =$

If A, B and C are interior angles of a triangle ABC, then Sin ($\frac{B+C}{2}\backslash frac\{B\; +\; C\}\{2\}$ ) =

$\frac{1-Ta{n}^{2}45^{\circ }}{1+Ta{n}^{2}45^{\circ }}\backslash frac\{1-Tan^2\; 45^\backslash circ\}\{1+Tan^2\; 45^\backslash circ\}$ can not be expressed as ………