Please solve it fast......

Dear Student ,

Please find below the required solution.

let's first calculate AB.
$$\begin{gathered} A{B}^{2 }= A{C}^2 + B{C}^2 [ pythagoras theorm ] \\ A{B}^{2 } = {12}^2 + 5^2 \\ A{B}^2 = 144 + 25 = 169 \\ AB = \sqrt{169} = 13 \\ 1) SinA = \frac{side oppoiste to angle A}{Hypoteneus} = \frac{5}{13} \\ CosA = \frac{Adjacent side to Angle A}{Hypoteneus} = \frac{12}{13} \\ \Rightarrow Si{n}^{2}A +\hspace{0.17em}Co{s}^{2}A = {\left(\frac{5}{13}\right)}^2+{\left(\frac{12}{13}\right)}^2= \frac{5^2 + {12}^2}{{13}^2}=\frac{25 +\hspace{0.17em}144}{169} \\ = \frac{169}{169} = 1 \\ Si{n}^{2}A +\hspace{0.17em}Co{s}^{2}A = 1 \\ 2) SecA = \frac{Hypoteneus}{Adjacent side to Angle A} = \frac{13}{12} \\ \tan A = \frac{side oppoiste to angle A}{Adjacent side to Angle A} = \frac{5}{12} \\ Se{c}^{2}A - Ta{n}^{2}A = {\left(\frac{13}{12}\right)}^2 - {\left(\frac{5}{12}\right)}^2 = \frac{{13}^2 - 5^2 }{{12}^2}= \frac{169 -25}{144} \\ = 144/144 = 1 \\ hence Se{c}^{2}A - Ta{n}^{2}A = 1 \end{gathered}$$
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