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Sine and Cosine Formulae and Their Applications — MATHS STD 11 Science — Question

Gujarat 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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