If $\tan \theta+3 \cot \theta-2 \sqrt{3}=0,0^{\circ}<\theta<90^{\circ}$, then what is the value of $\left({cosec}^2 \theta+\cos ^2 \theta\right) ?$
Answer & explanation
Correct answer: option 2
tan θ + 3cot θ - 2√3 = 0
tan θ + 3 × \(\frac{1}{tan θ}\) - 2√3 = 0
tan² θ - 2√3tan θ + 3 = 0
tan² θ - √3tan θ - √3tan θ + 3 = 0
tan θ ( tan θ - √3 ) - 1 ( tan θ - √3 ) = 0
( tan θ - 1 ). ( tan θ - √3 ) = 0
Either ( tan θ - 1 ) = 0 Or ( tan θ - √3 ) = 0
tan θ - 1 = 0 is not possible because 0º < θ < 90º
So, tan θ - √3 = 0
tan θ = √3
{ we know, tan60º = √3 }
Now,
( cosec² θ + cot² θ )
= ( cosec² 60º + cos² 60º )
= \(\frac{4}{3}\) + \(\frac{1}{4}\)
= \(\frac{19}{12}\)