If $3 \sec ^2 \theta+\tan \theta=7,0^{\circ}<\theta<90^{\circ}$, then the value of $\frac{{cosec} 2 \theta+\cos \theta}{\sin 2 \theta+\cot \theta}$ is:
Answer & explanation
Correct answer: option 3
3 sec²θ + tanθ = 7
{ we know, sec²θ - tan²θ = 1 }
3 ( 1 + tan²θ ) + tanθ = 7
3 tan²θ + tanθ - 4 = 0
3 tan²θ +4 tanθ - 3tanθ - 4 = 0
tanθ ( 3tanθ + 4 ) - 1 ( 3tanθ + 4 ) = 0
( tanθ - 1 ). ( 3tanθ + 4 ) = 0
Either ( tanθ - 1 ) = 0 Or ( 3tanθ + 4 ) = 0
( 3tanθ + 4 ) = 0 is not possible.
So, ( tanθ - 1 ) = 0
tanθ = 1
{ tan45º = 1 }
Now,
\(\frac{ cosec 2θ + cosθ}{ sin2θ + cotθ}\)
= \(\frac{ cosec 90º + cos45º}{ sin90º + cot45º}\)
= \(\frac{ 1 +1/√2}{ 1 +1 }\)
= \(\frac{ √2 +1}{ 2√2 }\)
= \(\frac{ 2 +√2}{ 4 }\)