====== 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. 

=====Question 7(i)=====
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*}

=====Question 7(ii)=====
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*}
 
=====Question 7(iii)=====
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*}
GOOD






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