Question 7, Review Exercise

Solutions of Question 7 of Review Exercise 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.

Show that: $\dfrac{4 \sin ^{2} \theta \cos \theta}{\cos 3 \theta+\cos \theta}=\tan 2 \theta \tan \theta$

Solution.

$$\begin{align*} LHS&=\frac{4 \sin^2 \theta \cos \theta}{\cos 3 \theta + \cos \theta}\\ &=\frac{4 \sin \theta\sin \theta \cos \theta}{4\cos^ 3 \theta-3\cos \theta + \cos \theta}\\ &=\frac{2 \sin2 \theta \sin \theta}{4\cos^ 3 \theta-2\cos \theta }\\ &=\frac{2 \sin2 \theta \sin \theta}{2\cos \theta(2\cos^ 2 \theta-1 )}\\ &=\frac{ \sin2 \theta \sin \theta}{\cos \theta\cos2\theta }\\ & = \tan 2 \theta \tan \theta\\ &=RHS \end{align*}$$

Show that: $\dfrac{\sin 10 \theta-\sin 4 \theta}{\sin 4 \theta+\sin 2 \theta}=\cos 7 \theta \sec \theta$

Solution.

$$\begin{align*} LHS&=\frac{\sin 10 \theta - \sin 4 \theta}{\sin 4 \theta + \sin 2 \theta} \\ &= \frac{2 \cos \left( \frac{10 \theta + 4 \theta}{2} \right) \sin \left( \frac{10 \theta - 4 \theta}{2} \right)}{2 \sin \left( \frac{4 \theta + 2 \theta}{2} \right) \cos \left( \frac{4 \theta - 2 \theta}{2} \right)} \\ &= \frac{2 \cos 7 \theta \sin 3 \theta}{2 \sin 3 \theta \cos \theta} \\ &= \cos 7 \theta \sec \theta\\ &=RHS \end{align*}$$

Show that: If $\sin (\alpha-\theta)=\cos (\alpha+\theta)$ prove that $\tan \alpha=1$.

Solution.

Given: \begin{align*} & \sin (\alpha - \theta) = \cos (\alpha + \theta) \\ \implies & \sin \alpha \cos \theta - \cos \alpha \sin \theta = \cos \alpha \cos \theta - \sin \alpha \sin \theta\end{align*} Dividing both sides by $\cos \alpha \cos \theta$ \begin{align*} & \frac{\sin \alpha \cos \theta}{\cos \alpha \cos \theta}-\frac{\cos \alpha \sin \theta}{\cos \alpha \cos \theta}=\frac{\cos \alpha \cos \theta}{\cos \alpha \cos \theta}+\frac{\sin \alpha \sin \theta}{\cos \alpha \cos \theta} \\ \implies & \tan \alpha -\tan \theta= 1-\tan \alpha \tan \theta\\ \implies & \tan \alpha +\tan \alpha \tan \theta =1+\tan \theta\\ \implies & \tan \alpha(1 + \tan \theta) =1+\tan \theta\\ \implies & \tan \alpha = \frac{1+ \tan \theta}{1+ \tan \theta}\\ \implies & \tan \alpha =1 \end{align*}

Alternative Method:
\begin{align*} & \sin (\alpha - \theta)= \cos (\alpha + \theta) \\ \implies &\sin \alpha \cos \theta - \cos \alpha \sin \theta = \cos \alpha \cos \theta - \sin \alpha \sin \theta \\ \implies &\sin \alpha \cos \theta + \sin \alpha \sin \theta = \cos \alpha \cos \theta + \cos \alpha \sin \theta \\ \implies &\sin \alpha (\cos \theta + \sin \theta) = \cos \alpha (\cos \theta + \sin \theta) \\ \implies & \frac{\sin \alpha}{\cos \alpha} = 1\\ \implies &\tan \alpha = 1 \end{align*}

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