Question 8(xvi, xvii & xviii) Exercise 8.2
Solutions of Question 8(xvi, xvii & xviii) 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.
Question 8(xvi)
Verify the identities: $\dfrac{1-\cos ^{2} \beta}{2-2 \cos \beta}=\cos ^{2} \dfrac{\beta}{2}$
Solution.
\begin{align*} LHS &= \dfrac{1-\cos ^{2} \beta}{2-2 \cos \beta}\\ &= \dfrac{\sin ^{2} \beta}{2-2 \cos \beta}\\ &=\dfrac{4\sin ^{2} \frac{\beta}{2} \cos ^{2} \frac{\beta}{2}}{2-2 (1-2\sin^2 \frac{\beta}{2} )}\quad \text{by using half angle identity}\\ &= \dfrac{4\sin ^{2} \frac{\beta}{2} \cos ^{2} \frac{\beta}{2}}{4\sin^2 \frac{\beta}{2} )}\\ &=\cos^2 \frac{\beta}{2}\\ &=RHS \end{align*}
Question 8(xvii)
Verify the identities: $\dfrac{\sin \theta}{1+\cos \theta}=\tan \dfrac{\theta}{2}$
Solution.
\begin{align*} LHS &= \dfrac{\sin \theta}{1+\cos \theta}\\ &= \dfrac{2\sin ^{2} \frac{\theta}{2} \cos \frac{\theta}{2}}{1+(1-2 \sin ^2\frac{\theta}{2})}\\ &=\dfrac{2\sin ^{2} \frac{\theta}{2} \cos \frac{\theta}{2}}{2(1- \sin ^2\frac{\theta}{2})}\\ &= \dfrac{2\sin ^{2} \frac{\theta}{2} \cos \frac{\theta}{2}}{2(\cos ^2\frac{\theta}{2})}\\ &=\tan \frac{\theta}{2}\\ &=RHS \end{align*}
Question 8(xviii)
Verify the identities: $\dfrac{1-\tan ^{2} \dfrac{x}{2}}{1+\tan ^{2} \dfrac{x}{2}}=\cos x$
Solution.
\begin{align*} LHS &=\dfrac{1-\tan ^{2} \dfrac{x}{2}}{1+\tan ^{2} \dfrac{x}{2}}\\ &= \dfrac{1- \dfrac{1-\cos x}{1+\cos x}}{1+\dfrac{1-\cos x}{1+\cos x}}\\ &=\dfrac{ \dfrac{1+\cos x-(1-\cos x)}{1+\cos x}}{\dfrac{1+\cos x+1-\cos x}{1+\cos x}}\\ &= \dfrac{2\cos x}{2}\\ &=\cos x\\ &=RHS \end{align*}
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