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Trigonometric Functions — Maths STD 11 Science — Question

CBSE BoardEnglish MediumSTD 11 ScienceMathsTrigonometric Functions5 Marks
Question
Prove that: $\frac{\sin\text{(A-B)}}{\cos\text{A}\cos\text{B}}+\frac{\sin\text{(B-C)}}{\cos\text{B}\cos\text{C}}+\frac{\sin\text{(C-A)}}{\cos\text{C}\cos\text{A}}=0$
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Answer

$\text{L.H.S}=\frac{\sin\text{(A-B)}}{\cos\text{A}\cos\text{B}}+\frac{\sin\text{(B-C)}}{\cos\text{B}\cos\text{C}}+\frac{\sin\text{(C-A)}}{\cos\text{C}\cos\text{A}}=0$
$=\frac{\sin\text{A}\cos\text{B}-\cos\text{A}\sin\text{B}}{\cos\text{A}\cos\text{B}}+\frac{\sin\text{B}\cos\text{C}-\cos\text{B}\sin\text{C}}{\cos\text{B}\cos\text{C}}\\\ \ +\frac{\sin\text{C}\cos\text{A}-\cos\text{C}\sin\text{A}}{\cos\text{C}\cos\text{A}}$
$=\frac{\sin\text{A}\cos\text{B}}{\cos\text{A}\cos\text{B}}-\frac{\cos\text{A}\sin\text{A}}{\cos\text{A}\cos\text{B}}+\frac{\sin\text{B}\cos\text{C}}{\cos\text{B}\cos\text{C}}\\\ \ -\frac{\cos\text{B}\sin\text{C}}{\cos\text{B}\cos\text{C}}+\frac{\sin\text{C}\cos\text{A}}{\cos\text{C}\cos\text{A}}-\frac{\cos\text{C}\sin\text{A}}{\cos\text{C}\cos\text{A}}$
$$$=\tan\text{A}-\tan\text{B}+\tan\text{B}-\tan\text{C}+\tan\text{C}-\tan\text{A}$
$=0$
$=\text{R.H.S}$
$\therefore\ \text{L.H.S}=\text{R.H.S}$
Hence proved.

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