Question
If $\sec A = \sqrt{2} ,$ find $:\frac{3 \cot ^2 A +2 \sin ^2 A }{\tan ^2 A -\cos ^2 A }.$
✓
Answer
$\sec A=\frac{\sqrt{2}}{1}$
i.e. $\frac{\text { hypotenuse }}{\text { base }}=\frac{ AC }{ AB }=\frac{\sqrt{2}}{1}$
Therefore if length of base $= x ,$ length of hypotenuse $=\sqrt{2} x$
Since
$AB^2 + BC^2 = AC^2\dots ...[$Using Pythagoras Theorem$]$
$(\sqrt{2} x)^2-(x)^2= BC ^2$
$BC ^2=2 x^2-x^2$
$BC ^2= x ^2$
$\therefore BC = x$
Now
$\cos A =\frac{1}{\sec A }=\frac{1}{\sqrt{2}}$
$\sin A =\frac{ BC }{ AC }=\frac{1}{\sqrt{2}}$
$\tan A =\frac{ BC }{ AB }=1$
$\cot A=\frac{1}{\tan A}=1$
Therefore
$\frac{3 \cot ^2 A +2 \sin ^2 A }{\tan ^2 A -\cos ^2 A }=\frac{3(1)^2+2\left(\frac{1}{\sqrt{2}}\right)^2}{1^2-\left(\frac{1}{\sqrt{2}}\right)^2}$
$=\frac{3+1}{1-\frac{1}{2}}$
$=\frac{4}{\frac{1}{2}}$
$=4 \times \frac{2}{1}$
$=8$
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