====== Question 3(vi, vii, viii, ix & x) Exercise 8.3 ======
Solutions of Question 3(vi, vii, viii, ix & x) of Exercise 8.3 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. 

=====Questio 3(vi)=====
Prove the identity $2\tan y \cos 3y= \sec y(\sin 4y-\sin 2y)$

** Solution. **

\begin{align*}
LHS & = 2\tan y \cos 3y \\
& = 2 \cdot \frac{\sin y}{\cos y} \cos 3y \\
& = \sec y (2 \cos 3y \sin y) \\
& = \sec y \left(\sin (3y+y)-\sin (3y-y) \right) \\
& = \sec y(\sin 4y - \sin 2y)\\
&= RHS
\end{align*}
GOOD
=====Questio 3(vii)=====
Prove the identity $\dfrac{ \sin 6 \beta + \sin 4 \beta}{\sin 6 \beta - \sin 4 \beta}=\tan 5 \beta \cot \beta$

** Solution. **

\begin{align*}
LHS & = \dfrac{ \sin 6 \beta + \sin 4 \beta}{\sin 6 \beta - \sin 4 \beta} \\
& = \dfrac{2 \sin \left( \frac{6\beta + 4\beta}{2} \right) \cos \left( \frac{6\beta - 4\beta}{2} \right)}{2 \cos \left( \frac{6\beta + 4\beta}{2} \right) \sin \left( \frac{6\beta - 4\beta}{2} \right)} \\
& = \dfrac{\sin 5\beta \cos 5\beta}{\sin \beta \cos \beta} \\
& = \tan 5\beta \cot \beta\\
&= RHS
\end{align*}
GOOD
=====Questio 3(viii)=====
Prove the identity $\dfrac{ \cot 3 \theta + \cot \theta}{\cot 3 \theta - \cot \theta}=-\cos 2 \theta \cot \theta$. m(

** Solution. **

\begin{align*}
LHS & = \dfrac{ \cot 3 \theta + \cot \theta}{\cot 3 \theta - \cot \theta} \\
& = \dfrac{\frac{\cos 3\theta}{\sin 3\theta} + \frac{\cos \theta}{\sin \theta}}{\frac{\cos 3\theta}{\sin 3\theta} - \frac{\cos \theta}{\sin \theta}} \\
& = \dfrac{\frac{\cos 3\theta \sin \theta + \sin 3\theta \cos \theta}{\sin 3\theta \sin \theta}}{\frac{\cos 3\theta \sin \theta - \sin 3\theta \cos \theta }{\sin 3\theta \sin \theta}} \\
& = \dfrac{ \sin 3\theta \cos \theta +\cos 3\theta \sin \theta }{-(\sin 3\theta \cos \theta-\cos 3\theta \sin \theta)} \\
& = \dfrac{\sin (3\theta+\theta)}{-\sin (3\theta-\theta)} \\
& = -\dfrac{\sin 4\theta}{\sin 2\theta} \\
& = -\dfrac{2\sin 2\theta \cos 2\theta}{\sin 2\theta} \\
& = -2 \cos 2\theta \\
&= RHS
\end{align*}
GOOD :!:

**Correction:**\\
The corrected version of the question can be stated as follows:
$$\dfrac{ \cot 3 \theta + \cot \theta}{\cot 3 \theta - \cot \theta}=-2\cos 2 \theta$$



=====Questio 3(ix)=====
Prove the identity $\dfrac{\cos 6x+\cos8x}{\sin6x-\sin4x}=\cot x \cos7x \sec 5x$

** Solution. **

\begin{align*}
LHS & = \dfrac{\cos 6x + \cos 8x}{\sin 6x - \sin 4x} \\
& = \dfrac{2 \cos \left( \frac{6x + 8x}{2} \right) \cos \left( \frac{6x - 8x}{2} \right)}{2 \cos \left( \frac{6x + 4x}{2} \right) \sin \left( \frac{6x - 4x}{2} \right)} \\
& = \dfrac{\cos(7x) \cos(-x)}{\cos(5x) \sin(x)} \\
& = \cos 7x  \dfrac{1}{\cos 5x} \dfrac{\cos x }{\sin x} \\
& = \cos 7x \sec 5x \cot x\\
&=RHS
\end{align*}
GOOD

=====Questio 3(x)=====
Prove the identity $\dfrac{\cos 2\alpha-\cos4 \alpha}{\sin2\alpha+\sin4\alpha}=\tan \alpha$

** Solution. **

\begin{align*}
LHS & = \dfrac{\cos 2\alpha - \cos 4\alpha}{\sin 2\alpha + \sin 4\alpha} \\
& = \dfrac{-2 \sin\left(\frac{2\alpha + 4\alpha}{2}\right) \sin\left(\frac{2\alpha - 4\alpha}{2}\right)}{2 \sin\left(\frac{2\alpha + 4\alpha}{2}\right) \cos\left(\frac{2\alpha - 4\alpha}{2}\right)} \\
& = \dfrac{-\sin 3\alpha  \sin(-\alpha)}{\sin 3\alpha  \cos \alpha } \\
& = \dfrac{\sin \alpha }{\cos \alpha} \\
& = \tan \alpha\\
&=RHS
\end{align*}
GOOD






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