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8.    If 3\cot A=4, check wether \frac{1-\tan ^{2}A}{1+\tan ^{2}A}=\cos ^{2}A-\sin ^{2}A or not.

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Given that,
3\cot A=4, 
\Rightarrow \cot = \frac{4}{3} = \frac{base}{perp.}
ABC is a right-angled triangle in which \angle B =90^0 and the length of the base AB is 4 units and length of perpendicular is 3 units

By using Pythagoras theorem,
In triangle ABC,
\\AC^2=AB^2+BC^2\\ AC = \sqrt{16+9}\\ AC = \sqrt{25}
AC = 5 units

So, 
\tan A = \frac{BC}{AB} = \frac{3}{4}
\cos A = \frac{AB}{AC} = \frac{4}{5}
\sin A = \frac{BC}{AC} = \frac{3}{5}
 \frac{1-\tan ^{2}A}{1+\tan ^{2}A}=\cos ^{2}A-\sin ^{2}A
Put the values of above trigonometric ratios, we get;
\Rightarrow \frac{1-9/4}{1+9/4} = \frac{16}{25}-\frac{9}{25}
\Rightarrow -\frac{5}{13} \neq \frac{7}{25}
LHS \neq RHS

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manish

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