A) 1
B) 0
C) -1
D) 2
Correct Answer: B
Solution :
Given, \[\cos ecA-\cot A=q\] \[={{\left( \cos ecA-\cot A \right)}^{2}}={{q}^{2}}\] \[=\cos e{{c}^{2}}A+{{\cot }^{2}}A-2\cos ecA\,.\,cot\,A={{q}^{2}}\] \[\therefore \] Put the value of \[{{q}^{2}}\]in \[\frac{{{q}^{2}}-1}{{{q}^{2}}+1}+\cos \,A\]. \[=\frac{\cos e{{c}^{2}}A+{{\cot }^{2}}A-2\cos ecA\,.\,\cot A-1}{\cos e{{c}^{2}}A+{{\cot }^{2}}A-2\cos ecA\,.\,\cot \,A+1}+\cos A\]\[=\frac{2\,{{\cot }^{2}}A-2\cos ec\,A\,.\,\cot \,A}{2\,\cos e{{c}^{2}}A-2\cos ec\,A\,.\,\cot \,A}+\cos A\] \[=\frac{2\cot A}{2\cos ecA}\left[ \frac{\cot A-\cos ecA}{\cos ec\,A-\cot \,A} \right]+\cos A\] \[=\cos \,A\left( -1 \right)+\cos A=0\]
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