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Trigonometric Functions — MATHS STD 11 Science — Question

Rajasthan BoardEnglish MediumSTD 11 ScienceMATHSTrigonometric Functions5 Marks
Question
$\frac{\cos^2\text{B}-\cos^2\text{C}}{\text{b + c}}+\frac{\cos^2\text{C}-\cos^2\text{A}}{\text{c + a}}+\frac{\cos^2\text{A}-\cos^2\text{B}}{\text{a + b}}=0$
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Answer

$\frac{\cos^2\text{B}-\cos^2\text{C}}{\text{b + c}}+\frac{\cos^2\text{C}-\cos^2\text{A}}{\text{c + a}}+\frac{\cos^2\text{A}-\cos^2\text{B}}{\text{a + b}}=0$
$\text{LHS}=\frac{\cos^2\text{B}-\cos^2\text{C}}{\text{b + c}}+\frac{\cos^2\text{C}-\cos^2\text{A}}{\text{c + a}}+\frac{\cos^2\text{A}-\cos^2\text{B}}{\text{a + b}}$
$=\frac{\cos^2\text{B}-\cos^2\text{C}}{\text{b + c}}+\frac{\cos^2\text{C}-\cos^2\text{A}}{\text{c + a}}+\frac{\cos^2\text{A}-\cos^2\text{B}}{\text{a + b}}$
$=\frac{1-\sin^2\text{B}-1+\sin^2\text{C}}{\text{b + c}}+\frac{1-\sin^2\text{C}-1+\sin^2\text{A}}{\text{c + a}}+\frac{1-\sin^2\text{A}-1+\sin^2\text{B}}{\text{a +b}}$
$=\frac{\sin^2\text{C}-\sin^2\text{B}}{\text{b + c}}+\frac{\sin^2\text{A}-\sin^2\text{C}}{\text{c + a}}+\frac{\sin^2\text{B}-\sin^2\text{A}}{\text{a + b}}$
$=\frac{\text{k}^2\text{c}^2-\text{k}^2\text{b}^2}{\text{b + c}}+\frac{\text{k}^2\text{a}^2-\text{k}^2\text{c}^2}{\text{c + a}}+\frac{\text{k}^2\text{b}^2-\text{k}^2\text{a}^2}{\text{a + b}}$
$=\text{k}^2\Big(\frac{\text{c}^2-\text{b}^2}{\text{b + c}}+\frac{\text{a}^2-\text{c}^2}{\text{c + a}}+\frac{\text{b}^2-\text{a}^2}{\text{a + b}}\Big)$
$=\text{k}^2(\text{c}-\text{b + a}-\text{c + b}-\text{a})$ $[\text{Using }\text{b}^2 -\text{a}^2 = (\text{b}-\text{a})(\text{b + a})]$
$=0=\text{RHS}$
Hence Proved

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