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Sine And Cosine Formulae And Their Applications — Maths STD 11 Science — Question

Maharashtra BoardEnglish MediumSTD 11 ScienceMathsSine And Cosine Formulae And Their Applications3 Marks
Question
$(\text{c}^2-\text{a}^2+\text{b}^2)\tan\text{A}=(\text{a}^2-\text{b}^2+\text{c}^2)\tan\text{B}=(\text{b}^2-\text{c}^2+\text{a}^2)\tan\text{C}$
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Answer

For any $\triangle\text{ABC},$ we have
$\cos\text{A}=\frac{\text{b}^2+\text{c}^2-\text{a}^2}{2\text{bc}}$
$\cos\text{B}=\frac{\text{a}^2+\text{c}^2-\text{b}^2}{2\text{ac}}$
$\cos\text{C}=\frac{\text{a}^2+\text{b}^2-\text{c}^2}{2\text{ab}}$
therefore,
$(\text{c}^2+\text{b}^2-\text{a}^2)\tan\text{A}=(\text{c}^2+\text{b}^2-\text{a}^2)\frac{\sin\text{A}}{\cos\text{A}}$
$=(\text{c}^2+\text{b}^2-\text{a}^2)\frac{\text{ka}}{\frac{\text{b}^2+\text{c}^2-\text{a}^2}{2\text{bc}}}$
$=2\text{kabc}$
Also,
$(\text{a}^2+\text{c}^2-\text{b}^2)\tan\text{B}=(\text{a}^2+\text{c}^2-\text{b})^2\frac{\sin\text{B}}{\cos\text{B}}$
$=(\text{a}^2+\text{c}^2-\text{b}^2)\frac{\text{kb}}{\frac{\text{a}^2+\text{c}^2-\text{b}^2}{2\text{ac}}}$
$=2\text{kabc}$
Now,
$(\text{a}^2+\text{b}^2-\text{c}^2)\tan\text{C}=(\text{a}^2+\text{b}^2-\text{c}^2)\frac{\sin\text{C}}{\cos\text{C}}$
$=(\text{a}^2+\text{b}^2-\text{c}^2)\frac{\text{kc}}{\frac{\text{a}^2+\text{b}^2-\text{c}^2}{2\text{ab}}}$
$=2\text{kabc}$
Hence proved.

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