====== 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*}
====Go to ====
[[math-11-nbf:sol:unit08:ex8-2-p10|< Question 8(xiii, xiv & xv) ]]
[[math-11-nbf:sol:unit08:ex8-2-p12|Question 8(xix, xx, xxi & xxii) >]]