Solutions of Question 8(xix, xx, xxi & xxii) of Exercise 8.2 of Unit 08: Fundamental of Trigonometry. This is unit of Model Textbook of Mathematics for Class XI published by National Book Foundation (NBF) as Federal Textbook Board, Islamabad, Pakistan.
Verify the identity: $$\frac{\sin 2 \alpha}{\sin \alpha}-\frac{\cos 2 \alpha}{\cos \alpha}=\sec \alpha$$
Solution.
\begin{align*}
LHS &= \dfrac{\sin 2 \alpha}{\sin \alpha}-\frac{\cos 2 \alpha}{\cos \alpha}\\
&= \dfrac{\sin 2 \alpha \cos\alpha - \cos 2 \alpha \sin \alpha}{\sin \alpha \cos\alpha}\\
&= \dfrac{\sin (2 \alpha - \alpha)}{\sin \alpha \cos\alpha}\\
&= \dfrac{\sin (\alpha)}{\sin \alpha \cos\alpha}\\
&= \dfrac{1}{\cos\alpha}\\
& = \sec\alpha \\
&=RHS
\end{align*}
Verify the identity: $2 \sin ^{2} \frac{\beta}{2}+\cos \beta=1$
Solution.
\begin{align*}
LHS & = 2 \sin ^{2} \frac{\beta}{2}+\cos \beta \\
& = 2 \sin^2 \frac{\beta}{2}+\cos^2 \frac{\beta}{2} - \sin^2 \frac{\beta}{2}\\
& = \sin^2 \frac{\beta}{2}+\cos^2 \frac{\beta}{2} \\
& = 1 \\
& = RHS
\end{align*}
Verify the identity: $$2 \cos y \sec 2 y=\frac{1}{\cos y-\sin y}+\frac{1}{\cos y+\sin y}$$
Solution.
\begin{align*}
RHS & = \frac{1}{\cos y-\sin y}+\frac{1}{\cos y+\sin y} \\
& = \frac{\cos y+\sin y+\cos y-\sin y}{(\cos y-\sin y)(\cos y+\sin y)} \\
& = \frac{2\cos y}{\cos^2 y-\sin^2 y} \\
& = \frac{2\cos y}{\cos 2y} \\
& = 2\cos y \sec 2y \\
& = LHS
\end{align*}
Verify the identity: $$2 \sin y \sec 2 y=\frac{1}{\cos y-\sin y}-\frac{1}{\cos y+\sin y}$$
Solution.
\begin{align*}
RHS & = \frac{1}{\cos y-\sin y}-\frac{1}{\cos y+\sin y} \\
& = \frac{\cos y+\sin y-\cos y+\sin y}{(\cos y-\sin y)(\cos y+\sin y)} \\
& = \frac{2\sin y}{\cos^2 y-\sin^2 y} \\
& = \frac{2\sin y}{\cos 2y} \\
& = 2\sin y \sec 2y \\
& = LHS
\end{align*}