Question
Short-Answer Questions:If $\cot\text{A}=\frac{4}{5}$ prove that $\frac{(\sin\text{A}+\cos\text{A})}{(\sin\text{A}-\cos\text{A})}=9$
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Answer
Given: $\cot\text{A}=\frac{4}{5}$Writing $\cot\text{A}=\frac{\cos\text{A}}{\sin\text{A}}$ and squaring the equation, we get:
$\frac{\cos^2\text{A}}{\sin^2\text{A}}=\frac{16}{25}$
$\Rightarrow25\cos^2\text{A}=16\sin^2\text{A}$
$\Rightarrow25\cos^2\text{A}=16-16\cos^2\text{A}$
$\Rightarrow\cos^2\text{A}=\frac{16}{41}$
$\Rightarrow\cos\text{A}=\frac{4}{\sqrt{41}}$
$\therefore\sin^2\text{A}=1-\cos^2\text{A}$
$1-\frac{16}{41}$
Now, $\sin\text{A}=\sqrt{\frac{25}{41}}$
$\Rightarrow\sin\text{A}=\frac{5}{\sqrt{41}}$
$\therefore\text{LHS}=\frac{\sin\text{A}+\cos\text{A}}{\sin\text{A}-\cos\text{A}}$
$=\frac{\frac{5}{\sqrt{41}}+\frac{4}{\sqrt{41}}}{\frac{5}{\sqrt{41}}-\frac{4}{\sqrt{41}}}$
$=\frac{9}{1}$
$=9=\text{RHS}$
$\frac{\cos^2\text{A}}{\sin^2\text{A}}=\frac{16}{25}$
$\Rightarrow25\cos^2\text{A}=16\sin^2\text{A}$
$\Rightarrow25\cos^2\text{A}=16-16\cos^2\text{A}$
$\Rightarrow\cos^2\text{A}=\frac{16}{41}$
$\Rightarrow\cos\text{A}=\frac{4}{\sqrt{41}}$
$\therefore\sin^2\text{A}=1-\cos^2\text{A}$
$1-\frac{16}{41}$
Now, $\sin\text{A}=\sqrt{\frac{25}{41}}$
$\Rightarrow\sin\text{A}=\frac{5}{\sqrt{41}}$
$\therefore\text{LHS}=\frac{\sin\text{A}+\cos\text{A}}{\sin\text{A}-\cos\text{A}}$
$=\frac{\frac{5}{\sqrt{41}}+\frac{4}{\sqrt{41}}}{\frac{5}{\sqrt{41}}-\frac{4}{\sqrt{41}}}$
$=\frac{9}{1}$
$=9=\text{RHS}$
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