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*}
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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