Question
If $3\cot\text{A}=4$ check whether $\frac{1+\tan^2\text{A}}{1+\tan^2\text{A}}=\cos^2\text{A}-\sin^2\text{A}$ or not.
✓
Answer
We have, $3\cot\text{A}=4\Rightarrow\cot\text{A}=\frac{4}{3}\Rightarrow\tan\text{A}=\frac{3}{4}$
In $\triangle\text{ABC},$ $\Rightarrow\text{AC}^2=(3)^2+(4)^2$ $\Rightarrow\text{AC}^2=25$ $\Rightarrow\text{AC}=5$ Now, $\frac{1-\tan^2\text{A}}{1+\tan^2\text{A}}=\cos^2\text{A}-\sin^2\text{A}$ $\Rightarrow\frac{1-\Big(\frac{3}{4}\Big)^2}{1+\Big(\frac{3}{4}\Big)^2}=\Big(\frac{4}{5}\Big)^2-\Big(\frac{3}{5}\Big)^2$ $\Rightarrow\frac{1-\frac{9}{16}}{1+\frac{9}{16}}=\frac{16}{25}-\frac{9}{25}$ $\Rightarrow\frac{7}{25}\times\frac{16}{16}=\frac{7}{25}$ $\Rightarrow\text{L.H.S}=\text{R.H.S}$
In $\triangle\text{ABC},$ $\Rightarrow\text{AC}^2=(3)^2+(4)^2$ $\Rightarrow\text{AC}^2=25$ $\Rightarrow\text{AC}=5$ Now, $\frac{1-\tan^2\text{A}}{1+\tan^2\text{A}}=\cos^2\text{A}-\sin^2\text{A}$ $\Rightarrow\frac{1-\Big(\frac{3}{4}\Big)^2}{1+\Big(\frac{3}{4}\Big)^2}=\Big(\frac{4}{5}\Big)^2-\Big(\frac{3}{5}\Big)^2$ $\Rightarrow\frac{1-\frac{9}{16}}{1+\frac{9}{16}}=\frac{16}{25}-\frac{9}{25}$ $\Rightarrow\frac{7}{25}\times\frac{16}{16}=\frac{7}{25}$ $\Rightarrow\text{L.H.S}=\text{R.H.S}$
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