By using Pythagoras theorem, it is known that \(a^2+b^2=c^2\). Divide the two sides of the equation by \(a^2\), \(b^2\) and \(c^2\), we get:
\(\dfrac{a^2}{a^2}+\dfrac{b^2}{a^2}=\dfrac{c^2}{a^2}\)
\(1+\left( \dfrac{b}{a} \right)^2=\left( \dfrac{c}{a} \right)^2\)
\(1+\cot^2{A}=\cosec^2{A}\)
\(\dfrac{a^2}{b^2}+\dfrac{b^2}{b^2}=\dfrac{c^2}{b^2}\)
\(\left( \dfrac{a}{b} \right)^2+1=\left( \dfrac{c}{b} \right)^2\)
\(1+\tan^2{A}=\sec^2{A}\)
\(\dfrac{a^2}{c^2}+\dfrac{b^2}{c^2}=\dfrac{c^2}{c^2}\)
\(\left( \dfrac{a}{c} \right)^2+\left( \dfrac{b}{c} \right)^2=1\)
\(\sin^2{A}+\cos^2{A}=1\)
Without using a calculator, find the value of each of the following:
(a)
\(\text{sin}^2 \ A + \text{cos}^2 \ A=1\\ \text{sin}^2 (-430^{\circ}) + \text{cos}^2(-430^{\circ}) =1.\)
(b)
\(\begin{aligned} &1+ \text{tan}^2 \ A = \text{sec}^2 \ A\\\\ &\text{tan}^2 \begin{pmatrix} \dfrac{\pi}{3} \end{pmatrix} - \text{sec}^2\begin{pmatrix} \dfrac{\pi}{3} \end{pmatrix} =-1 .\end{aligned}\)
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