\begin{array}{l}{\frac{1}{s c c \theta-\tan \theta}+\frac{1}{\sec \theta+\tan \theta}=\frac{2}{\cos \theta}} \\ {=\frac{\operatorname{scc} \theta+\tan \theta+\sec \theta-\tan \theta}{(\sec \theta-\tan \theta)(\sec \theta+\tan \theta)}} \\ {=\frac{2 \sec \theta}{\operatorname{scc}^{2} \theta-\tan ^{2} \theta}} \\ {=\frac{2 \sec \theta}{1+\tan ^{2} \theta-\tan ^{2} \theta}} \\ {=2 \sec \theta}\end{array}

\begin{array}{l}{\frac{1}{s c c \theta-\tan \theta}+\frac{1}{\sec \the

1.	\begin{array}{l}{\frac{1}{s c c \theta-\tan \theta}+\frac{1}{\sec \theta+\tan \theta}=\frac{2}{\cos \theta}} \\ {=\frac{\operatorname{scc} \theta+\tan \theta+\sec \theta-\tan \theta}{(\sec \theta-\tan \theta)(\sec \theta+\tan \theta)}} \\ {=\frac{2 \sec \theta}{\operatorname{scc}^{2} \theta-\tan ^{2} \theta}} \\ {=\frac{2 \sec \theta}{1+\tan ^{2} \theta-\tan ^{2} \theta}} \\ {=2 \sec \theta}\end{array}
Answer» ssc result 10th board