If $\cot \theta+\tan \theta=2 \sec \theta, 0^{\circ}<\theta<90^{\circ}$, then the value of $\frac{\tan 2 \theta-\sec \theta}{\cot 2 \theta+{cosec} \theta}$.
Answer & explanation
Correct answer: option 3
cot θ + tan θ = 2 sec θ
\(\frac{cosθ}{sinθ}\) + \(\frac{sinθ}{cosθ}\) = 2 . \(\frac{1}{cosθ}\)
\(\frac{cos²θ + sin²θ}{sinθcosθ}\) = 2 . \(\frac{1}{cosθ}\)
{ cos²θ + sin²θ = 1 }
\(\frac{1}{sinθ}\) = 2
sinθ = \(\frac{1}{2}\)
{ we know, sin 30º = \(\frac{1}{2}\) }
So, θ = 30º
Now,
\(\frac{tan2θ - secθ}{cot2θ + cosecθ}\)
= \(\frac{tan60º - sec30º}{cot60º + cosec30º}\)
= \(\frac{ √3 - 2/√3 }{1/√3 + 2}\)
= \(\frac{ 1 }{ 2√3 + 1 }\)
= \(\frac{ 1 }{ 2√3 + 1 }\) × \(\frac{ 2√3 - 1 }{ 2√3 - 1 }\)
= \(\frac{ 2√3 - 1 }{ 11}\)