If \(\int \frac{d \theta}{\cos ^2 \theta(\tan 2 \theta+\sec 2 \theta)}=\lambda \tan \theta+2 \log |f(\theta)|+c\)
where \(\mathrm{C}\) is a constant of integration, then the ordered pair \((\lambda, f(\theta))\) is equal to

\(\int \frac{d \theta}{\cos ^2 \theta(\tan 2 \theta+\sec 2 \theta)}=\int \frac{\sec ^2 \theta d \theta}{\frac{\sin 2 \theta+1}{\cos 2 \theta}} \)
\( =\int \frac{\left(\cos ^2 \theta-\sin ^2 \theta\right) \sec ^2 \theta d \theta}{2 \sin \theta \cdot \cos \theta+\sin ^2 \theta+\cos ^2 \theta} \)
\( =\int \frac{(\cos \theta+\sin \theta)(\cos \theta-\sin \theta) \sec ^2 \theta d \theta}{(\cos \theta+\sin \theta)^2} \)
\( =\int \frac{\cos \theta-\sin \theta}{\cos \theta+\sin \theta} \sec ^2 \theta d \theta\)
\(=\int \frac{1-\tan \theta}{1+\tan \theta} \sec ^2 \theta d \theta \quad \quad \text { [dividing } \mathrm{N}^{\mathrm{r}}=\) \(\text { and } \Delta^{\mathrm{r}} \text { by } \cos \theta \text { ] } \)
\( =\int \frac{1-t}{1+t} d t \quad \quad \text { [let } \tan \theta=t \text { ] } \)
\( =\int\left(-1+\frac{2}{1+t}\right) d t \)
\( =-t+2 \log |1+t|+C \)
\( =-\tan \theta+2 \log |1+\tan \theta|+C \)
\( \text { Comparing we get } \lambda=-1 \text { and } f(\theta)=1+\tan \theta\)