CBSE BoardEnglish MediumSTD 11 ScienceMathsTrigonometric Functions1 Mark
Question
If $3\sin\text{x}+5\cos\text{x}=5,$ then write the value of $5\sin\text{x}-3\cos\text{x}.$
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Answer
$3\sin\text{x}+5\cos\text{x}=5$ (Given)
Squaring both the side:
$9\sin^2\text{x}+25\cos^2\text{x}+30\sin\text{x}\cos\text{x}=25$
$30\sin\text{x}\cos\text{x}=25-9\sin^2\text{x}-25\cos^2\text{x}\cdots(1)$
We have to find the value of $5\sin\text{x}-3\cos\text{x}.$
$(5\sin\text{x}-3\cos\text{x})^2=25\sin^2\text{x}+9\cos^2\text{x}-30\sin\text{x}\cos\text{x}$
$(5\sin\text{x}-3\cos\text{x})^2=25\sin^2\text{x}+6\cos^2\text{x}-(25-9\sin^2\text{x}-25\cos^2\text{x})$ [From (1)]
$(5\sin\text{x}-3\cos\text{x})^2=34\sin^2\text{x}+34\cos^2\text{x}-25$
$(5\sin\text{x}-3\cos\text{x})^2=34-25$ $(\because\sin^2\text{x}+\cos^2\text{x}=1)$
$(5\sin\text{x}-3\cos\text{x})^2=9$
$5\sin\text{x}-3\cos\text{x}=\pm3$
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