The value of \(\rm \frac{(\sec θ+\tan θ)(1-\sin θ)}{\cot θ(1+\tan θ+\sec θ)(1+\cot θ-cosec θ)}\) 0° < θ < 90° is equal to:

Formula used:

sec θ = 1/cos θ

tan θ = sin θ/cos θ

cot θ = cos θ/sin θ

cosec θ = 1/sin θ

sin² θ + cos² θ = 1

Given:

\(\rm \frac{(\sec θ+\tan θ)(1-\sin θ)}{\cot θ(1+\tan θ+\sec θ)(1+\cot θ-cosec θ)}\)

⇒ \(\rm \frac{(\frac{1}{\cos θ}+\frac{\sin θ}{\cos θ})(1-\sin θ)}{\frac{\cos θ}{\sin θ}(1+\frac{\sin θ}{\cos θ}+\frac{1}{\cos θ})(1+\frac{\cos θ}{\sin θ}-\frac{1}{\sin θ})}\)

⇒  \(\rm \frac{(\frac{1+\sin θ}{\cos θ})(1-\sin θ)}{\frac{\cos θ}{\sin θ}(\frac{\cos θ+\sin θ+1}{\cos θ})(\frac{\sin θ+\cos θ-1}{\sin θ})}\)

⇒ \(\rm \frac{1-\sin^2 θ}{\cos θ} \times \frac{1}{(\frac{\cos θ+\sin θ+1}{\sin θ})(\frac{\sin θ+\cos θ-1}{\sin θ})}\)

⇒ \(\rm \frac{\cos^2 θ}{\cos θ} \times \frac{\sin^2 θ}{(\cos θ+\sin θ)^2-1}\)

⇒ \(\rm \cos θ \times \frac{\sin^2 θ}{\cos^2 θ+2\sin θ\cos θ+\sin^2 θ-1}\)

⇒ \(\rm \cos θ \times \frac{\sin^2 θ}{2\sin θ\cos θ}\) = \(\rm \frac{\sin θ}{2}\)

∴ The value of the expression is 1/2sin θ.