Question

√1+sinA/1-sinA=tan (π/4 + A/2)​

Answer 1

TO PROVE :–

$\\ \implies \bf \sqrt{ \dfrac{1 + \sin(A) }{1 - \sin(A)}} = \tan \left(\dfrac{\pi}{4} + \dfrac{A}{2} \right)$

SOLUTION :–

• Let's take L.H.S. –

$\\ \: \: = \: \: \bf \sqrt{ \dfrac{1 + \sin(A) }{1 - \sin(A)}}$

• Now rationalization –

$\\ \: \: = \: \: \bf \sqrt{ \dfrac{1 + \sin(A) }{1 - \sin(A)} \times \dfrac{1 + \sin(A) }{1 + \sin(A)}}$

$\\ \: \: = \: \: \bf \sqrt{ \dfrac{ \{1 + \sin(A) \}\{1 + \sin(A) \}}{ \{1 - \sin(A) \}\{1 + \sin(A) \}}}$

$\\ \: \: = \: \: \bf \sqrt{ \dfrac{ \{1 + \sin(A) \}^{2}}{ \{1 - \sin^{2} (A) \}}}$

$\\ \: \: = \: \: \bf \sqrt{ \dfrac{ \{1 + \sin(A) \}^{2}}{\cos^{2} (A)}}$

$\\ \: \: = \: \: \bf\dfrac{1 + \sin(A)}{\cos(A)}$

• We should write this as –

$\\ \: \: = \: \: \bf\dfrac{1 + \sin(\pi/2 - A)}{\cos(\pi/2 - A)}$

$\\ \: \: = \: \: \bf\dfrac{1 + \cos(\pi/2 - A)}{\sin(\pi/2 - A)}$

$\\ \: \: = \: \: \bf\dfrac{2 \cos ^{2} (\pi/4 - A/2)}{2\sin(\pi/4 - A/2)\cos(\pi/4 - A/2)}$

$\\ \: \: = \: \: \bf\dfrac{ \cos (\pi/4 - A/2)}{\sin(\pi/4 - A/2)}$

$\\ \: \: = \: \: \bf \cot (\pi/4 - A/2)$

$\\ \: \: = \: \: \bf \tan(\pi/2 - \{\pi/4 - A/2 \})$

$\\ \: \: = \: \: \bf \tan(\pi/2 -\pi/4 + A/2)$

$\\ \: \: = \: \: \bf \tan \{(2\pi–\pi)/4 + A/2 \}$

$\\ \: \: = \: \: \bf \tan \left(\dfrac{\pi}{4} + \dfrac{A}{2} \right)$

$\\ \: \: = \: \: \bf R.H.S.$

$\\ \implies \red{\underbrace{\bf Hence \: \: Proved}}$