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?

-\(\frac{1}{8}\)

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°

∴ \(\frac{tan  θ - sec  θ}{2 tan  θ + 3 sec  θ}\) 

= \(\frac{\frac{1}{sqrt{3}}\; - \; \frac{2}{sqrt{3}}}{\frac{2}{sqrt{3}}\; + \; \frac{3\times 2}{sqrt{3}}}\) 

=  \(\frac{\frac{1}{sqrt{3}}}{\frac{8}{sqrt{3}}}\) 

= -\(\frac{1}{8}\)