A) \[p-q=s\left( \frac{\tan \alpha -\tan 3\alpha }{\tan 3\alpha \tan \alpha } \right)\]
B) \[p-q=s\left( \frac{\tan 3\alpha -\tan \alpha }{3\tan 3\alpha tan\alpha } \right)\]
C) \[p-q=s\left( \frac{\tan 3\alpha -\tan \alpha }{\tan 3\alpha \tan \alpha } \right)\]
D) \[p-q=s\left( \frac{\tan 2\alpha }{\tan 3\alpha \tan \alpha } \right)\]
Correct Answer: C
Solution :
In \[\Delta ABC,\] |
\[\tan \alpha =\frac{s}{p}\]\[\Rightarrow \]\[p=\frac{s}{\tan \alpha }\] ?(i) |
In \[\Delta BDC,\] |
\[\tan \beta =\frac{s}{q}\] |
\[\Rightarrow \] \[q=\frac{s}{\tan 3\alpha }\] ?(ii) |
On subtracting Eq. (ii) from Eq. (i), we get |
\[p-q=\frac{s}{\tan \alpha }-\frac{s}{\tan 3\alpha }\] |
\[=s\left( \frac{\tan 3\alpha -tan\alpha }{\tan \alpha \tan 3\alpha } \right)\] |
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