Question

prove it
$\frac{1 + \sin \theta - \cos \theta}{1 + \sin \theta + \cos \theta} = \tan \frac{1}{2} \theta$

Answer 1

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$\bold{\underline{Question-}}$

prove it
$\frac{1 + \sin \theta - \cos \theta}{1 + \sin \theta + \cos \theta} = \tan \frac{1}{2} \theta$




$\bold{\underline{Answer-}}$

LHS = $\frac{1 + \sin \theta - \cos \theta}{1 + \sin \theta + \cos \theta}$



$= \frac{2 {sin}^{2} \frac{ \theta}{2} + 2sin \frac{ \theta}{2} cos \frac{ \theta}{2} }{2 {cos}^{2} \frac{ \theta}{2} + 2sin \frac{ \theta}{2} cos \frac{ \theta}{2} }$



$= \frac{2sin \frac{ \theta}{2}(sin \frac{ \theta}{2} + cos \: \frac{ \theta}{2} ) }{2cos \frac{ \theta}{2} (cos \frac{ \theta}{2} + sin \frac{ \theta}{2} ) }$


$= \frac{sin \frac{ \theta}{2} }{cos \frac{ \theta}{2} }$


$= tan \frac{1}{2} \theta$


LHS = RHS

Hence proved,

Answer 2

Answer:

Step-by-step explanation:

LHS =  \frac{1 + \sin \theta - \cos \theta}{1 + \sin \theta + \cos \theta}  

=  \frac{2 {sin}^{2} \frac{ \theta}{2}   + 2sin \frac{ \theta}{2} cos \frac{ \theta}{2} }{2 {cos}^{2} \frac{ \theta}{2} + 2sin \frac{ \theta}{2} cos \frac{ \theta}{2}   }  \\  

=  \frac{2sin \frac{ \theta}{2}(sin \frac{ \theta}{2}  + cos \:  \frac{ \theta}{2} ) }{2cos \frac{ \theta}{2} (cos \frac{ \theta}{2} + sin \frac{ \theta}{2} ) }  \\  

=  \frac{sin \frac{ \theta}{2} }{cos \frac{ \theta}{2} }  

= tan \frac{1}{2}  \theta

LHS = RHS  

Hence proved,