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.

Verify the identities: $\csc 2 \alpha-\cot 2 \alpha=\tan \alpha$

Solution.

$$\begin{align*} LHS &= \csc 2 \alpha-\cot 2 \alpha\\ &=\frac{1}{\sin 2 \alpha}- \frac{\cos2 \alpha}{\sin 2\alpha }\\ &=\frac{1-\cos2 \alpha}{\sin2 \alpha}\\ &= \frac{2\sin^2 \alpha}{2\sin \alpha \cos \alpha}\\ &=\tan \alpha\\ &=RHS \end{align*}$$

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.

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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