If 3 \sin ^{2} \theta-\cos \theta-1=0, 0^{\circ}

SSC CGL 18 Aug 2021 Shift 2 Paper

Question : 52 of 100

Marks: +1, -0

3 \sin ^{2} θ-\cos θ-1=0

0^{∘} < θ < 90^{∘}

, then what is the value of

\cot θ+cosec θ

Solution:

Given,

3 \sin ^{2} θ-\cos θ-1=0

⇒ 3(1-\cos ^{2} θ)-\cos θ-1=0

⇒ \;\; 3-3 \cos ^{2} θ-\cos θ-1=0

⇒ \;\; 3 \cos ^{2} θ+\cos θ-2=0

⇒ 3 \cos ^{2} θ+3 \cos θ-2 \cos θ-2=0

⇒ 3 \cos θ(\cos θ+1)-2(\cos θ+1)=0

⇒ \;\; (3 \cos θ-2)(\cos θ+1)=0

⇒ \;\; \cos θ=\;{2}/{3}, \cos θ=-1

⇒ \;\; \cos θ=\;{2}/{3}=\;{\;\text " Base "\;}/{\;\text " Hypotenuse "\;}

By Pythagoras theorem,

(A C)^{2} \;\; =(A B)^{2}+(B C)^{2}

⇒ \;\; (3)^{2} \;\; =P^{2}+(2)^{2}

⇒ \;\; 9-4 \;\; =P^{2}

⇒ \;\; P^{2} \;\; =5

⇒ \;\; P \;\; =√{5}

⇒ \;\; \sin θ \;\; =\;{P}/{H}=\;{√{5}}/{3}

∴ \;\; \cot θ+ cosec θ

\;\; =\;{\cos θ}/{\sin θ}+\;{1}/{\sin θ}

\;\; =\;{2}/{√{5}}+\;{1}/{√{5}}

\;\; =\;{2}/{√{5}}+\;{3}/{√{5}}=\;{5}/{√{5}}

\;\; =\;{√{5} × √{5}}/{√{5}}=√{5}

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