If 12 \cot ^{2} \\theta-31 {\cosec} \\theta+32=0, 0^\circ

SSC CGL 6 06 2019 Shift 2 Paper

Question : 73 of 100

Marks: +1, -0

12 \cot ^{2} \θ-31 \ {\cosec} \θ+32=0,

0° < \θ < 90°,

thenthe values of tan

\θ

will be:

Solution:

12 \cot ^{2} \θ-31 \ {\cosec} \θ+32=0

12 \cos ^{2} \θ-31 \sin \θ+32 \sin ^{2} \θ=0

12-12 \sin ^{2} \θ-31 \sin \θ+32 \sin ^{2} \θ=0

20 \sin ^{2} \θ-31 \sin \θ+12=0

20 \sin ^{2} \θ-16 \sin \θ-15 \sin \θ+12=0

4 \sin \θ(5 \sin \θ-4)-3(5 \sin \θ-4)=0

\sin \θ=\ {4}/{5}, \ {3}/{4}

when

\sin \θ=\ {4}/{5}

\tan \θ=\ {4}/{3}

when

\sin \θ=\ {3}/{4}

\tan \θ=\ {3}/{\√{7}}

\tan \θ=\ {3}/{\√{7}}

\ {3 \√{7}}/{7}

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