====== Question 8(xiii, xiv & xv)  Exercise 8.2 ======
Solutions of Question 8(xiii, xiv & xv) 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(xiii)=====
Verify the identities: $\csc 2 \alpha-\cot 2 \alpha=\tan \alpha$

** Solution. **

\begin{align*}
LHS &= \csc 2 \alpha-\cot 2 \alpha\\
&=\frac{1}{\sin 2 \alpha}- \frac{\cos2 \alpha}{\sin 2\alpha }\\
&=\frac{1-\cos2 \alpha}{\sin2 \alpha}\\
&= \frac{2\sin^2 \alpha}{2\sin \alpha \cos \alpha}\\
&=\tan \alpha\\
&=RHS
\end{align*}
GOOD
=====Question 8(xiv)=====
Verify the identities: $\dfrac{\cos 3 x-\sin 3 x}{\cos x-\sin x}=\dfrac{2+\sin 2 x}{2}$

** Solution. **

\begin{align*}
LHS & =  \dfrac{\cos 3 x-\sin 3 x}{\cos x-\sin x} \\
& = \dfrac{(\cos 3 x-\sin 3 x)(\cos x + \sin x)}{(\cos x-\sin x)((\cos x + \sin x))} \\
&= \dfrac{\cos 3x \cos x +\cos 3x \sin x - \sin 3x\cos x -\sin 3x \sin x}{\cos^2 x-\sin^2 x} \\
&= \dfrac{(\cos 3x \cos x  -\sin 3x \sin x) +(\cos 3x \sin x - \sin 3x\cos x)}{\cos 2x} \\
&= \dfrac{\cos (3x+x)+\sin(3x-x)}{\cos 2x} \\
&= \dfrac{\cos 4x+\sin2x}{\cos 2x} \\
&\neq RHS \\
\end{align*}

This question doesn't seems correct.







=====Question 8(xv)=====
Verify the identities: $\dfrac{\sin 3 \alpha}{\sin \alpha}-\dfrac{\cos 3 \alpha}{\cos \alpha}=2$

** Solution. **

\begin{align*}
LHS &= \dfrac{\sin 3 \alpha}{\sin \alpha}-\dfrac{\cos 3 \alpha}{\cos \alpha}\\
&= \dfrac{\sin 3 \alpha \cos\alpha - \cos 3 \alpha \sin \alpha}{\sin \alpha \cos\alpha}\\
&= \dfrac{\sin (3 \alpha - \alpha)}{\sin \alpha \cos\alpha}\\
&= \dfrac{\sin (2 \alpha)}{\sin \alpha \cos\alpha}\\
&= \dfrac{2\sin\alpha \cos\alpha}{\sin \alpha \cos\alpha}\\
& = 2 =RHS
\end{align*}

**Alternative Method**
\begin{align*}
LHS &= \dfrac{\sin 3 \alpha}{\sin \alpha}-\dfrac{\cos 3 \alpha}{\cos \alpha}\\
&= \dfrac{3\sin \alpha-4\sin^3 \alpha}{\sin \alpha}-\dfrac{4\cos ^3 \alpha-3\cos \alpha}{\cos \alpha}\\
&=3-4\sin^2 \alpha-4\cos^2\alpha+3\\
&= 6-4 (\sin^2 \alpha+\cos^2\alpha)\\
&=2\\
&=RHS
\end{align*}









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