Question
If 3 cot A = 4, check whether $\frac { 1 - \tan ^ { 2 } A } { 1 + \tan ^ { 2 } A } = \cos ^ { 2 } A - \sin ^ { 2 } A$ or not.
✓
Answer
Give that 3cot A = 4
Or cot $A = \frac{4}{3}$
Consider a right angle triangle $\Delta ABC$ right angled at point B.
$\cot A = \frac{{Side\;adjacent\;to\;\angle A}}{{Side\;opposite\;to\;\angle A}}$
$\frac{{AB}}{{BC}} = \frac{4}{3}$
If AB is 4K, BC will be 3K. where K is a positive integer
Now in $\Delta ABC$
$(AC)^2 = (AB)^2 + (BC)^2$
$= (4K)^2 + (3K)^2$
$= 16 K^2 + 9K^2$
$= 25K^2$
$AC = 5K$
$\cos A = \frac{{Side\;adjacent\;to\;\angle A}}{{hypotenuse}} = \frac{{AB}}{{AC}}$
$ = \frac{{4K}}{{5K}} = \frac{4}{5}$
$\sin A = \frac{{Side\;opposite\;to\;\angle A}}{{hypotenuse}} = \frac{{BC}}{{AC}}$
$ = \frac{{3K}}{{5K}} = \frac{3}{5}$
$\tan A = \frac{{Side\;opposite\;to\;\angle A}}{{Side\;adjacent\;to\;angle\;A}} = \frac{{BC}}{{AB}}$
$ = \frac{{3K}}{{4K}} = \frac{3}{4}$
$\frac{{1 - {{\tan }^2}A}}{{1 + {{\tan }^2}A}} = \frac{{1 - {{\left( {\frac{3}{4}} \right)}^2}}}{{1 + {{\left( {\frac{3}{4}} \right)}^2}}} = \frac{{1 - \frac{9}{{16}}}}{{1 + \frac{9}{{16}}}}$
$ = \frac{{\frac{7}{{16}}}}{{\frac{{25}}{{16}}}} = \frac{7}{{25}}$
${\cos ^2}A - {\sin ^2}A = {\left( {\frac{4}{5}} \right)^2} - {\left( {\frac{3}{5}} \right)^2}$
$ = \frac{{16}}{{25}} - \frac{9}{{25}} = \frac{7}{{25}}$
Hence $\frac{{1 - {{\tan }^2}A}}{{1 + {{\tan }^2}A}} = {\cos ^2}A - {\sin ^2}A$
Or cot $A = \frac{4}{3}$
Consider a right angle triangle $\Delta ABC$ right angled at point B.
$\cot A = \frac{{Side\;adjacent\;to\;\angle A}}{{Side\;opposite\;to\;\angle A}}$
$\frac{{AB}}{{BC}} = \frac{4}{3}$
If AB is 4K, BC will be 3K. where K is a positive integer
Now in $\Delta ABC$
$(AC)^2 = (AB)^2 + (BC)^2$
$= (4K)^2 + (3K)^2$
$= 16 K^2 + 9K^2$
$= 25K^2$
$AC = 5K$
$\cos A = \frac{{Side\;adjacent\;to\;\angle A}}{{hypotenuse}} = \frac{{AB}}{{AC}}$
$ = \frac{{4K}}{{5K}} = \frac{4}{5}$
$\sin A = \frac{{Side\;opposite\;to\;\angle A}}{{hypotenuse}} = \frac{{BC}}{{AC}}$
$ = \frac{{3K}}{{5K}} = \frac{3}{5}$
$\tan A = \frac{{Side\;opposite\;to\;\angle A}}{{Side\;adjacent\;to\;angle\;A}} = \frac{{BC}}{{AB}}$
$ = \frac{{3K}}{{4K}} = \frac{3}{4}$
$\frac{{1 - {{\tan }^2}A}}{{1 + {{\tan }^2}A}} = \frac{{1 - {{\left( {\frac{3}{4}} \right)}^2}}}{{1 + {{\left( {\frac{3}{4}} \right)}^2}}} = \frac{{1 - \frac{9}{{16}}}}{{1 + \frac{9}{{16}}}}$
$ = \frac{{\frac{7}{{16}}}}{{\frac{{25}}{{16}}}} = \frac{7}{{25}}$
${\cos ^2}A - {\sin ^2}A = {\left( {\frac{4}{5}} \right)^2} - {\left( {\frac{3}{5}} \right)^2}$
$ = \frac{{16}}{{25}} - \frac{9}{{25}} = \frac{7}{{25}}$
Hence $\frac{{1 - {{\tan }^2}A}}{{1 + {{\tan }^2}A}} = {\cos ^2}A - {\sin ^2}A$
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