Question 3(vi, vii, viii, ix & x) Exercise 8.3

Solutions of Question 3(vi, vii, viii, ix & x) of Exercise 8.3 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.

Prove the identity $2\tan y \cos 3y= \sec y(\sin 4y-\sin 2y)$

Solution.

\begin{align*} LHS & = 2\tan y \cos 3y \\ & = 2 \cdot \frac{\sin y}{\cos y} \cos 3y \\ & = \sec y (2 \cos 3y \sin y) \\ & = \sec y \left(\sin (3y+y)-\sin (3y-y) \right) \\ & = \sec y(\sin 4y - \sin 2y)\\ &= RHS \end{align*}

Prove the identity $\dfrac{ \sin 6 \beta + \sin 4 \beta}{\sin 6 \beta - \sin 4 \beta}=\tan 5 \beta \cot \beta$

Solution.

\begin{align*} LHS & = \dfrac{ \sin 6 \beta + \sin 4 \beta}{\sin 6 \beta - \sin 4 \beta} \\ & = \dfrac{2 \sin \left( \frac{6\beta + 4\beta}{2} \right) \cos \left( \frac{6\beta - 4\beta}{2} \right)}{2 \cos \left( \frac{6\beta + 4\beta}{2} \right) \sin \left( \frac{6\beta - 4\beta}{2} \right)} \\ & = \dfrac{\sin 5\beta \cos 5\beta}{\sin \beta \cos \beta} \\ & = \tan 5\beta \cot \beta\\ &= RHS \end{align*}

Prove the identity $\dfrac{ \cot 3 \theta + \cot \theta}{\cot 3 \theta - \cot \theta}=-\cos 2 \theta \cot \theta$.

Solution.

\begin{align*} LHS & = \dfrac{ \cot 3 \theta + \cot \theta}{\cot 3 \theta - \cot \theta} \\ & = \dfrac{\frac{\cos 3\theta}{\sin 3\theta} + \frac{\cos \theta}{\sin \theta}}{\frac{\cos 3\theta}{\sin 3\theta} - \frac{\cos \theta}{\sin \theta}} \\ & = \dfrac{\frac{\cos 3\theta \sin \theta + \sin 3\theta \cos \theta}{\sin 3\theta \sin \theta}}{\frac{\cos 3\theta \sin \theta - \sin 3\theta \cos \theta }{\sin 3\theta \sin \theta}} \\ & = \dfrac{ \sin 3\theta \cos \theta +\cos 3\theta \sin \theta }{-(\sin 3\theta \cos \theta-\cos 3\theta \sin \theta)} \\ & = \dfrac{\sin (3\theta+\theta)}{-\sin (3\theta-\theta)} \\ & = -\dfrac{\sin 4\theta}{\sin 2\theta} \\ & = -\dfrac{2\sin 2\theta \cos 2\theta}{\sin 2\theta} \\ & = -2 \cos 2\theta \\ &= RHS \end{align*}

Correction:
The corrected version of the question can be stated as follows: $$\dfrac{ \cot 3 \theta + \cot \theta}{\cot 3 \theta - \cot \theta}=-2\cos 2 \theta$$

Prove the identity $\dfrac{\cos 6x+\cos8x}{\sin6x-\sin4x}=\cot x \cos7x \sec 5x$

Solution.

\begin{align*} LHS & = \dfrac{\cos 6x + \cos 8x}{\sin 6x - \sin 4x} \\ & = \dfrac{2 \cos \left( \frac{6x + 8x}{2} \right) \cos \left( \frac{6x - 8x}{2} \right)}{2 \cos \left( \frac{6x + 4x}{2} \right) \sin \left( \frac{6x - 4x}{2} \right)} \\ & = \dfrac{\cos(7x) \cos(-x)}{\cos(5x) \sin(x)} \\ & = \cos 7x \dfrac{1}{\cos 5x} \dfrac{\cos x }{\sin x} \\ & = \cos 7x \sec 5x \cot x\\ &=RHS \end{align*}

Prove the identity $\dfrac{\cos 2\alpha-\cos4 \alpha}{\sin2\alpha+\sin4\alpha}=\tan \alpha$

Solution.

\begin{align*} LHS & = \dfrac{\cos 2\alpha - \cos 4\alpha}{\sin 2\alpha + \sin 4\alpha} \\ & = \dfrac{-2 \sin\left(\frac{2\alpha + 4\alpha}{2}\right) \sin\left(\frac{2\alpha - 4\alpha}{2}\right)}{2 \sin\left(\frac{2\alpha + 4\alpha}{2}\right) \cos\left(\frac{2\alpha - 4\alpha}{2}\right)} \\ & = \dfrac{-\sin 3\alpha \sin(-\alpha)}{\sin 3\alpha \cos \alpha } \\ & = \dfrac{\sin \alpha }{\cos \alpha} \\ & = \tan \alpha\\ &=RHS \end{align*}

< Question 3(i, ii, iii, iv & v) Question 3(xi, xii & xiii)>