If \[\frac{{{\cos }^{2}}\theta }{{{\cot }^{2}}\theta -{{\cos }^{2}}\theta }=3\]and \[0{}^\circ <\theta <90{}^\circ ,\] then the value of \[\theta \] is [SSC (10+2) 2011] |
A) \[30{}^\circ \]
B) \[45{}^\circ \]
C) \[60{}^\circ \]
D) None of these
Correct Answer: C
Solution :
\[\frac{{{\cos }^{2}}\theta }{{{\cot }^{2}}\theta -{{\cos }^{2}}\theta }=3\] |
\[\Rightarrow \] \[{{\cos }^{2}}\theta =3{{\cot }^{2}}\theta -3{{\cos }^{2}}\theta \] |
\[\Rightarrow \] \[4{{\cos }^{2}}\theta =3{{\cot }^{2}}\theta \] |
\[\Rightarrow \] \[4{{\cos }^{2}}\theta -\frac{3{{\cos }^{2}}\theta }{{{\sin }^{2}}\theta }=0\] |
\[\Rightarrow \] \[{{\cos }^{2}}\theta \left( 4-\frac{3}{{{\sin }^{2}}\theta } \right)=0\] |
\[\Rightarrow \] \[4-\frac{3}{{{\sin }^{2}}\theta }=0\] |
\[\Rightarrow \] \[4{{\sin }^{2}}\theta -3=0\] |
\[\Rightarrow \] \[{{\sin }^{2}}\theta =\frac{3}{4}\] |
\[\Rightarrow \] \[\sin \theta =\frac{\sqrt{3}}{2}=\sin 60{}^\circ \] |
\[\therefore \] \[\theta =60{}^\circ \] |
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