Question 5 and 6, Review Exercise

Solutions of Question 5 and 6 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.

Find the values of $\tan \theta$ when $\tan \left(\theta-45^{\circ}\right)=\frac{1}{3}$.

Solution.

\begin{align*} & \frac{\tan \theta - \tan 45^{\circ}}{1 + \tan \theta \cdot \tan 45^{\circ}} =\frac{1}{3}\\ \implies & \frac{\tan \theta - 1}{1 + \tan \theta}= \frac{1}{3}\\ \implies & 3 \tan \theta - 3 = 1 + \tan \theta \\ \implies & 2 \tan \theta = 4 \\ \implies & \tan \theta = 2 \end{align*}

If $\sin (\alpha+\theta)=2 \cos (\alpha-\theta)$ prove that $\tan \alpha=\dfrac{2-\tan \theta}{1-2 \tan \theta}$.

Solution.

\begin{align*} & \sin (\alpha+\theta)=2 \cos (\alpha-\theta)\\ \implies & \sin \alpha \cos \theta+\cos \alpha \sin \theta =2(\cos \alpha \cos \theta+\sin \alpha \sin \theta)\\ \implies & \sin \alpha \cos \theta+\cos \alpha \sin \theta=2\cos \alpha \cos \theta+2\sin \alpha \sin \theta\end{align*} Divided 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{2\cos \alpha \cos \theta}{\cos \alpha \cos \theta}+\frac{2\sin \alpha \sin \theta}{\cos \alpha \cos \theta}\\ \implies & \tan \alpha +\tan \theta =2+2 \tan\alpha \tan \theta\\ \implies & \tan \alpha(1-2\tan \theta)=2-\tan \theta\\ \implies & \tan \alpha =\frac{2-\tan \theta}{1-2 \tan \theta} \end{align*}

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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