Question
If $\tan\text{A}=\text{n}\tan\text{B}$ and $\sin\text{A}=\text{m}\sin\text{B},$ prove that $\cos^2\text{A}=\frac{\big(\text{m}^2-1\big)}{\text{n}^2-1}.$
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Answer
We have, $\tan\text{A}=\text{n}\tan\text{B}$
$\Rightarrow\cot\text{B}=\frac{\text{n}}{\tan\text{A}}\dots(\text{i})$
Again, $\sin\text{A}=\text{m}\sin\text{B}$
$\Rightarrow\text{cosec B}=\frac{\text{m}}{\sin\text{A}}\dots(\text{ii})$
Squaring (i) and (ii) and subtracting (ii) from (i), we get:
$\frac{\text{m}^2}{\sin^2\text{A}}-\frac{\text{n}^2}{\tan^2\text{A}}=\text{cosec}^2\text{B}-\cot^2\text{B}$
$\Rightarrow\frac{\text{m}^2}{\sin^2\text{A}}-\frac{\text{n}^2\cos}{\sin^2\text{A}}=1$
$\Rightarrow\text{m}^2-\text{n}^2\cos^2\text{A}=\sin^2\text{A}$
$\Rightarrow\text{m}^2-\text{n}^2\cos^2\text{A}=1-\cos^2\text{A}$
$\Rightarrow\text{n}^2\cos^2\text{A}-\cos^2\text{A}=\text{m}^2-1$
$\Rightarrow\cos^2\text{A}\big(\text{n}^2-1\big)=\big(\text{m}^2-1\big)$
$\Rightarrow\cos^2\text{A}=\frac{\big(\text{m}^2-1\big)}{\big(\text{n}^2-1\big)}$
$\therefore\ \cos^2\text{A}=\frac{\big(\text{m}^2-1\big)}{\big(\text{n}^2-1\big)}$
$\Rightarrow\cot\text{B}=\frac{\text{n}}{\tan\text{A}}\dots(\text{i})$
Again, $\sin\text{A}=\text{m}\sin\text{B}$
$\Rightarrow\text{cosec B}=\frac{\text{m}}{\sin\text{A}}\dots(\text{ii})$
Squaring (i) and (ii) and subtracting (ii) from (i), we get:
$\frac{\text{m}^2}{\sin^2\text{A}}-\frac{\text{n}^2}{\tan^2\text{A}}=\text{cosec}^2\text{B}-\cot^2\text{B}$
$\Rightarrow\frac{\text{m}^2}{\sin^2\text{A}}-\frac{\text{n}^2\cos}{\sin^2\text{A}}=1$
$\Rightarrow\text{m}^2-\text{n}^2\cos^2\text{A}=\sin^2\text{A}$
$\Rightarrow\text{m}^2-\text{n}^2\cos^2\text{A}=1-\cos^2\text{A}$
$\Rightarrow\text{n}^2\cos^2\text{A}-\cos^2\text{A}=\text{m}^2-1$
$\Rightarrow\cos^2\text{A}\big(\text{n}^2-1\big)=\big(\text{m}^2-1\big)$
$\Rightarrow\cos^2\text{A}=\frac{\big(\text{m}^2-1\big)}{\big(\text{n}^2-1\big)}$
$\therefore\ \cos^2\text{A}=\frac{\big(\text{m}^2-1\big)}{\big(\text{n}^2-1\big)}$
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