Question

If $\sec A=\frac{17}{8}$, prove that $1-2{\sin }^{2}A=2{\cos }^2-1$

Answer 1

$\sec A=\frac{17}{8}\Rightarrow \cos A=\frac{8}{17}$

$\Rightarrow \sec A=\sqrt{1-{\cos }^{2}A}=\sqrt{1-\frac{64}{289}}=\sqrt{\frac{225}{289}}=\frac{15}{17}$

Now, $1-2{\sin }^{2}A=1-2{\left(\frac{15}{17}\right)}^2=1-\frac{2\times 225}{289}=\frac{289-450}{289}$

$\Rightarrow 1-2{\sin }^{2}A=\frac{-161}{289}$

$\&2{\cos }^{2}A-1=2\times {\left(\frac{8}{17}\right)}^2-1=\frac{128}{289}-1=\frac{128-289}{289}=\frac{-161}{289}$

$\Rightarrow 1-2{\sin }^{2}A=2{\cos }^{2}A-1$

Hence, the answer is proved.