Trigonometric Identities Question 315
Question: If $ \frac{3\pi }{4}<\alpha <\pi , $ then $ \sqrt{cose{c^{2}}\alpha +2\cot \alpha } $ is equal to
[Pb. CET 2000; AMU 2001; MP PET 2004]
Options:
A) $ 1+\cot \alpha $
B) $ 1-\cot \alpha $
C) $ -1-\cot \alpha $
D) $ -1+\cot \alpha $
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Answer:
Correct Answer: C
Solution:
$ \sqrt{cose{c^{2}}\alpha +2\cot \alpha }=\sqrt{1+{{\cot }^{2}}\alpha +2\cot \alpha }=|1+\cot \alpha | $ But $ \frac{3\pi }{4}<\alpha <\pi \Rightarrow \cot \alpha <-1\Rightarrow 1+\cot \alpha <0 $ Hence, $ |1+\cot \alpha |=-(1+\cot \alpha ) $ .
BETA
