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 tanθ when tan(θ45)=13.

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*} GOOD

If sin(α+θ)=2cos(αθ) prove that tanα=2tanθ12tanθ.

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 cosαcosθ \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*} GOOD

If sin(αθ)=cos(α+θ) prove that tanα=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 cosαcosθ \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*} GOOD