Prove that: (45^ +1 2 )= 1+ 1-

Question

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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