3 Marks Question

Question 13 Marks

Prove that:
$
\tan \left(45^{\circ}+\frac{1}{2}\right)=\sqrt{\frac{1+\sin \phi}{1-\sin \phi}}
$

Answer

$\begin{aligned}\text { L.H.S. }= & \tan \left(45^{\circ}+\frac{1}{2} \phi\right) \\& =\frac{\tan 45^{\circ}+\tan \frac{\phi}{2}}{1-\tan 45^{\circ} \tan \frac{\phi}{2}} \\& =\frac{1+\tan \phi / 2}{1-\tan \phi / 2}=\frac{1+\frac{\sin \phi / 2}{\cos \phi / 2}}{1-\frac{\sin \phi / 2}{\cos \phi / 2}} \\& =\frac{\cos \phi / 2+\sin \phi / 2}{\cos \phi / 2-\sin \phi / 2} \\& =\frac{(\cos \phi / 2+\sin \phi / 2)(\cos \phi / 2+\sin \phi / 2)}{(\cos \phi / 2-\sin \phi / 2)(\cos \phi / 2+\sin \phi / 2)} \\& =\frac{\cos \phi / 2+\sin ^2 \phi / 2+2 \sin \phi / 2 \cos \phi / 2}{\cos ^2 \phi / 2-\sin \phi / 2} \\& =\frac{1+\sin \phi}{\cos \phi} \\\text { R.H.S. } & =\sqrt{\frac{1+\sin \phi}{1-\sin \phi}}\end{aligned}$
Multiplying the numerator and denominator by
$\begin{aligned}\sqrt{1+\sin \phi} & \\& =\sqrt{\frac{1+\sin \phi}{1-\sin \phi}} \times \sqrt{\frac{1+\sin \phi}{1+\sin\phi}}=\sqrt{\frac{(1+\sin \phi)^2}{1-\sin ^2 \phi}} \\& =\sqrt{\frac{(1+\sin \phi)^2}{\cos ^2 \phi}}=\frac{1+\sin \phi}{\cos \phi}\end{aligned}$
$\text {L.H.S. = R.H.S.}$

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Question 23 Marks

If $\tan (A+B)=n \tan (A-B)$ show that :
$
(n+1) \sin 2 B=(n-1) \sin 2 A
$

Answer

Given, $\frac{\tan ( A + B )}{\tan ( A - B )}=\frac{n}{1}$
From Compondendo-Dividendo Rule
$\begin{array}{l}\Rightarrow \quad \frac{\tan (A+B)+\tan (A-B)}{\tan (A+B)-\tan (A-B)}=\frac{n+1}{n-1} \\\Rightarrow \quad \frac{\frac{\sin (A+B)}{\cos (A+B)}+\frac{\sin (A-B)}{\cos (A-B)}}{\frac{\sin (A+B)}{\cos (A+B)}-\frac{\sin (A-B)}{\cos (A-B)}}=\frac{n+1}{n-1}\end{array}$
$\Rightarrow \frac{\frac{\sin ( A + B ) \cos ( A - B )+\sin ( A - B ) \cos ( A + B )}{\cos ( A + B ) \cos ( A - B )}}{\frac{\sin ( A + B ) \cos ( A - B )-\sin ( A - B ) \cos ( A + B )}{\cos ( A + B ) \cos ( A - B )}}=\frac{n+1}{n-1}$
$\Rightarrow \quad \frac{\sin [( A + B )+( A - B )]}{\sin [( A + B )-( A - B )]}=\frac{n+1}{n-1}$
$\Rightarrow \quad \frac{\sin 2 A}{\sin 2 B}=\frac{n+1}{n-1}$
$\Rightarrow(n+1) \sin 2 A=(n-1) \sin 2 A$ Hence proved.

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Question 33 Marks

Prove that:
$4 \sin \alpha \cdot \sin \left(\alpha+\frac{\pi}{3}\right) \sin \left(\alpha+\frac{2 \pi}{3}\right)=\sin 3 \alpha$

Answer

$\begin{array}{l} \text { L.H.S. } 4 \sin \alpha \sin \left(\alpha+\frac{\pi}{3}\right) \sin \left(\alpha+\frac{2 \pi}{3}\right) \\ = 2 \sin \alpha\left[2 \sin \left(\alpha+\frac{\pi}{3}\right) \cdot \sin \left(\alpha+\frac{2 \pi}{3}\right)\right] \\ = 2 \sin \alpha\left[\cos \left(\alpha+\frac{\pi}{3}-\alpha-\frac{2 \pi}{3}\right)-\cos \left(\alpha+\frac{\pi}{3}+\alpha+\frac{2 \pi}{3}\right)\right] \\ {[\because 2 \sin A \sin B =\cos ( A - B )-\cos ( A + B )] } \\ = 2 \sin \alpha\left[\cos \left(-\frac{\pi}{3}\right)-\cos (\pi+2 \alpha)\right] \\ = 2 \sin \alpha\left[\cos \frac{\pi}{3}+\cos 2 \alpha\right] \quad \left[\begin{array}{c}\because \cos (\pi+\theta)=-\cos \theta \\ \text { and } \cos (-\theta)=\cos \theta\end{array}\right]\ \\ = 2 \sin \alpha\left[\frac{1}{2}+\cos 2 \alpha\right] \end{array}$
$\begin{array}{l}=\sin \alpha+2 \sin \alpha \cos 2 \alpha \\ =\sin \alpha+[\sin (\alpha+2 \alpha)-\sin (2 \alpha-\alpha)]\end{array}$
$[\because 2 \cos A \cos B=\sin (A+B)-\sin (A-B)]$
$=\sin \alpha+\sin 3 \alpha-\sin \alpha=\sin 3 \alpha=$ R.H.S.

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