====== 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. 


=====Question 5=====
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*}
GOOD
=====Question 6(i)=====
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*}
GOOD
=====Question 6(ii)=====
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*}
GOOD
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