If \[\theta \] is an acute angle and \[{{\tan }^{2}}\theta +\frac{1}{{{\tan }^{2}}\theta }=2,\] then the value of \[\theta \] is |
A) \[45{}^\circ \]
B) \[30{}^\circ \]
C) \[60{}^\circ \]
D) \[15{}^\circ \]
Correct Answer: A
Solution :
Given, \[{{\tan }^{2}}\theta +\frac{1}{{{\tan }^{2}}\theta }=2\]\[\Rightarrow \]\[\frac{{{\tan }^{4}}\theta +1}{{{\tan }^{2}}\theta }=2\] |
\[\Rightarrow \] \[{{\tan }^{2}}\theta -2{{\tan }^{2}}\theta +1=\theta \] |
Let \[y={{\tan }^{2}}\theta \] |
\[\therefore \] \[{{y}^{2}}-2y+1=0\] |
\[\Rightarrow \] \[{{y}^{2}}-y-y+1=0\] |
\[\Rightarrow \] \[y\,(y-1)-1\,(y-1)=0\] |
\[\Rightarrow \] \[(y-1)(y-1)=0\] |
\[\therefore \] \[y=1\] |
\[\because \] \[ta{{n}^{2}}\theta =1\] |
\[\Rightarrow \] \[\tan \theta =\pm \,1\]\[\Rightarrow \]\[\theta =45{}^\circ ,\]\[135{}^\circ \] |
But \[\theta \] is an acute angle, therefore \[\theta =45{}^\circ \] |
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