If cot θ + tan θ = 2 sec θ; where 0 < θ < 90°, then the value of \(\frac{tan θ - sec θ}{2 tan θ + 3 sec θ}\) is?
Answer & explanation
Correct answer: option 2
cot θ + tan θ = 2 sec θ ⇒ \(\frac{cos θ}{sin θ}\) + \(\frac{sin θ}{cos θ}\) = \(\frac{2}{cos θ}\)
⇒ \(\frac{cos^2 θ + sin^2 θ}{sin θ. cos θ}\) = \(\frac{2}{cos θ}\) ⇒ cosec θ = 2
or θ = 30°
∴ \(\frac{tan θ - sec θ}{2 tan θ + 3 sec θ}\)
= \(\frac{\frac{1}{sqrt{3}}\; - \; \frac{2}{sqrt{3}}}{\frac{2}{sqrt{3}}\; + \; \frac{3\times 2}{sqrt{3}}}\)
= \(\frac{\frac{1}{sqrt{3}}}{\frac{8}{sqrt{3}}}\)
= -\(\frac{1}{8}\)