Trigonometric Ratios Of Multiple And Sub Multiple Angles — Maths STD 11 Science — Question
CBSE BoardEnglish MediumSTD 11 ScienceMathsTrigonometric Ratios Of Multiple And Sub Multiple Angles1 Mark
Question
If $\text{a}=\text{b}\cos\frac{2\pi}{3}=\text{c}\cos\frac{4\pi}{3},$ then write the value of ab + bc + ca.
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Answer
Let, $\text{a}=\text{b}\cos\frac{2\pi}{3}=\text{c}\cos\frac{4\pi}{3}=\text{k}$
$\Rightarrow\text{a}=\text{k},\text{b}=\frac{\text{k}}{\cos\frac{2\pi}{3}}$ and $\text{c}=\frac{\text{k}}{\cos\frac{4\pi}{3}}$
Now, ab + bc + ca $=\text{k}\times\frac{\text{k}}{\cos\frac{2\pi}{3}}+\frac{\text{k}}{\cos\frac{2\pi}{3}}\times\frac{\text{k}}{\cos\frac{4\pi}{3}}+\frac{\text{k}}{\cos\frac{4\pi}{3}}\times\text{k}$
$=\text{k}^2\sec\frac{2\pi}{3}+\text{k}^2\sec\frac{2\pi}{3}\sec\frac{4\pi}{3}+\text{k}^2\sec\frac{4\pi}{3}$
$=\text{k}^2\sec\Big(\frac{\pi}{2}+\frac\pi6\Big)+\text{k}^2\sec\Big(\frac{\pi}{2}+\frac\pi6\Big)\sec\Big(\pi+\frac{\pi}{3}\Big)+\text{k}^2\sec\Big(\pi+\frac{\pi}{3}\Big)$
$=-\text{k}^2\text{cosec}\frac{\pi}{6}+\text{k}^2\text{cosec}\frac{\pi}{6}\sec\frac{\pi}{3}-\text{k}^2\sec\frac{\pi}{3}$
$=-\text{k}^2\times2+\text{k}^2\times2\times2-\text{k}^2\times2$
$=-4\text{k}^2+4\text{k}^2=0$
$\therefore\text{ ab + bc + ca}=0$
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