Question

If sec A = $\frac{17}{8}$, verify that $\frac{3-4{\sin }^{2}A}{4{\cos }^{2}A-3}=\frac{3-{\tan }^{2}A}{1-3{\tan }^{2}A}$

Answer 1

Given $\sec A=\frac{17}{8}$

$\cos A=\frac{1}{\sec A}=\frac{8}{17}$

Also, $\cos A=\frac{adjacentside}{Hypotenuse}$

$adjacentside=8$ , $Hypotenuse=17$

From Pythagoras theorem,

$(Hypotenuse{)}^2=(oppositeside{)}^2+(adjacentside{)}^2$

${17}^2=(oppositeside{)}^2+8^2$

$(oppositeside{)}^2=289-64=225$

$oppositeside=15$

$\sin A=\frac{oppositeside}{Hypotenuse}=\frac{15}{17}$

$\tan A=\frac{oppositeside}{adjacentside}=\frac{15}{8}$

To verify, $\frac{3-4{\sin }^{2}A}{4{\cos }^{2}A-3}=\frac{3-{\tan }^{2}A}{1-3{\tan }^{2}A}$

$\frac{3-4{\left(\frac{15}{17}\right)}^2}{4{\left(\frac{8}{17}\right)}^2-3}=\frac{3-{\left(\frac{15}{8}\right)}^2}{1-3{\left(\frac{15}{8}\right)}^2}$

$\frac{3-\left(\frac{900}{289}\right)}{\left(\frac{256}{289}\right)-3}=\frac{3-\left(\frac{225}{64}\right)}{1-\left(\frac{675}{64}\right)}$

$\frac{\left(\frac{867-900}{289}\right)}{\left(\frac{256-867}{289}\right)}=\frac{\left(\frac{192-225}{64}\right)}{\left(\frac{64-675}{64}\right)}$

$\frac{\left(\frac{-33}{289}\right)}{\left(\frac{-611}{289}\right)}=\frac{\left(\frac{-33}{64}\right)}{\left(\frac{-611}{64}\right)}$

$\frac{33}{611}=\frac{33}{611}$

$\therefore$ $\frac{3-4{\sin }^{2}A}{4{\cos }^{2}A-3}=\frac{3-{\tan }^{2}A}{1-3{\tan }^{2}A}$

Hence Verified