In a $Δ A B C,$ which of the following quantities denote the area of the triangle?

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$(\frac{a^{2} - b^{2}}{2}) (\frac{\sin A \sin B}{\sin (A - B)})$

$\frac{r_{1} r_{2} r_{3}}{\sqrt{\sum r_{1} r_{2}}}$

$\frac{a^{2} + b^{2} + c^{2}}{\cot A + \cot B + \cot C}$

$r^{2} \cot \frac{A}{2} . \cot \frac{B}{2} . \cot \frac{C}{2}$

answer is A, B, D.

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Detailed Solution

$A) At θ = \frac{π}{6}, x = \log [\cot (\frac{π}{4} + θ)]$
$\begin{array}{l} x = \log [\cot (\frac{π}{4} + \frac{π}{6})] \\ e^{x} = \frac{\cot \frac{π}{4} \cot \frac{π}{6} - 1}{\cot \frac{π}{6} + \cot \frac{π}{4}} \\ e^{x} = \frac{\sqrt{3} - 1}{\sqrt{3} + 1} \Rightarrow e^{- x} = \frac{\sqrt{3} + 1}{\sqrt{3} - 1} \\ \sinh (x) = \frac{e^{x} - e^{- x}}{2} = \frac{\frac{\sqrt{3} - 1}{\sqrt{3} + 1} - \frac{\sqrt{3} + 1}{\sqrt{3} - 1}}{2} \\ = \frac{(3 + 1 - 2 \sqrt{3}) - (3 + 1 + 2 \sqrt{3})}{2 \times 2} \\ = \frac{- 4 \sqrt{3}}{4} = - \sqrt{3} \end{array}$
B)(ii) for $x \in (\frac{- π}{2}, \frac{π}{2})$
$\begin{array}{l} logsec x = y \\ \sec x = e^{y} \\ \cos x = e^{- y} \\ \frac{\sinh y}{\cosh y} = \frac{e^{y} - e^{- y}}{e^{y} + e^{- y}} \\ \frac{\cosh y - \sinh y}{\cosh y + \sinh y} = \frac{e^{- y}}{e^{y}} = \cos^{2} x \\ \frac{\cosh \frac{y}{2} - \sinh \frac{y}{2}}{\cosh \frac{y}{2} + \sinh \frac{y}{2}} = \cos x \end{array}$
$\begin{array}{l} \frac{2 \cosh \frac{y}{2}}{2 \sinh \frac{y}{2}} = \frac{1 + \cos x}{1 - \cos x} = \cot^{2} \frac{x}{2} \\ \coth \frac{y}{1} = \cot^{2} \frac{x}{2} = \cos e c^{2} \frac{x}{2} - 1 \\ \frac{y}{2} = \coth^{- 1} (\cos e c^{2} \frac{x}{2} - 1) \\ y = 2 \coth^{- 1} (\cos e c^{2} \frac{x}{2} - 1) \end{array}$
C(iii) $\sec h^{- 1} x = \log_{e} (\frac{1 + \sqrt{1 - x^{2}}}{x})$
$\begin{array}{l} \tanh^{- 1} x = \frac{1}{2} \log (\frac{1 + x}{1 - x}) \\ \sinh^{- 1} x = \log (x + \sqrt{x^{2} + 1}) \\ e^{\sec h^{- 1} \frac{1}{2}} = e^{\log_{e} (\frac{1 + \sqrt{1 - \frac{1}{4}}}{\frac{1}{2}})} \\ e^{\log_{e} (\frac{1 + \frac{\sqrt{3}}{2}}{\frac{1}{2}})} = e^{\log_{e}^{(2 + \sqrt{3})}} = 2 + \sqrt{3} \\ e^{\tanh^{- 1} \frac{1}{2}} = e^{\frac{1}{2} \log (\frac{\frac{3}{2}}{\frac{1}{2}})} = \sqrt{3} \\ e^{\sinh^{- 1} \frac{1}{2}} = e^{\log (\frac{1}{2} + \sqrt{\frac{1}{4} + 1})} \\ e^{\log_{e} (\frac{1}{2} + \frac{\sqrt{5}}{2})} = \frac{\sqrt{5} + 1}{2} \\ e^{(\sec h^{- 1} \frac{1}{2} + \tanh^{- 1} \frac{1}{2} + \sinh^{- 1} \frac{1}{2})} = (2 + \sqrt{3}) (\sqrt{3}) \frac{(\sqrt{5} + 1)}{2} \\ = \frac{(2 \sqrt{3} + 3) (\sqrt{5} + 1)}{2} \\ = \frac{2 \sqrt{15} + 2 \sqrt{3} + 3 \sqrt{5} + 3}{2} \\ = \frac{3 + 3 \sqrt{5} + 2 \sqrt{3} + 2 \sqrt{15}}{2} \end{array}$
D(IV) $f (x) = \log \sqrt{\frac{1 + x}{1 - x}} + \log (\frac{1 + \sqrt{1 - x^{2}}}{x}) + \coth^{- 1} x + \log (\frac{1 + \sqrt{1 + x^{2}}}{x}) + \sinh^{- 1} x$ $N o w \log \sqrt{\frac{1 + x}{1 - x}}$ will be defined when
$\begin{array}{l} \sqrt{\frac{1 + x}{1 - x}} > 0 = \frac{1 + x}{1 - x} > 0 \\ \Rightarrow x \in (- 1, 1) \\ \log (\frac{1 + \sqrt{1 - x^{2}}}{x}) w i l l b e d i f i n e d w h e n \\ \frac{1 + \sqrt{1 - x^{2}}}{x} > 0 a n d 1 - x^{2} > 0 a n d x \neq 0 \\ 1 - x^{2} > 0, x \neq 0, x > 0 \\ (1 - x) (1 + x) > 0, x \neq 0, x > 0 \\ \Rightarrow x \in (0, 1) \\ \log (\frac{1 + \sqrt{1 + x^{2}}}{x}) w i l l b e d e f i n e d w h e n \frac{1 + \sqrt{1 + x^{2}}}{x} > 0 \\ \Rightarrow x > 0 \\ ∴ x \in (0, \infty) \\ d o m a i n o f \coth^{- 1} x i s R - [- 1, 1] & d o m a i n o f \sinh^{- 1} x i s R \\ ∴ f (x) w i l l b e d e f i n e d o n (0, 1) \end{array}$