Dasukede kudasai
Trigonometry prove
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Answer:
Step-by-step explanation:
Formulas used:
[tex]sin( A +B) = 2sin Acos B\\\\sin A= sin (\frac{A}{2} + \frac{A}{2}) = 2 \ sin \frac{A}{2}cos \frac{A}{2}\\\\cos(A + B) = cosA cosB - sinA sinB\\\\cos A = cos(\frac{A}{2} + \frac{A}{2}) = cos \frac{A}{2} cos \frac{A}{2} - sin \frac{A}{2} sin \frac{A}{2}[/tex]
[tex]= cos^2 \frac{A}{2} - sin^2 \frac{A}{2}[/tex]
[tex]1 - sin^2 \frac{A}{2} = cos^2 \frac{A}{2}[/tex]
[tex]\frac{sin\frac{A}{2}}{cos\frac{A}{2}} = tan\frac{A}{2}[/tex]
Given :
LHS =
[tex]\frac{sin \frac{A}{2} + sin A }{1 + cos \frac{A}{2} + cosA}\\\\=\frac{sin \frac{A}{2} + 2sin \frac{A}{2} cos \frac{A}{2} }{1 + cos \frac{A}{2} +cos ^2\frac{A}{2} -sin^2 \frac{A}{2} }\\\\= \frac{sin\frac{A}{2}( 1 +2cos \frac{A}{2} ) }{cos \frac{A}{2} + cos ^2 \frac{A}{2} + 1 - sin^2\frac{A}{2} }\\\\= \frac{sin\frac{A}{2}( 1 +2cos \frac{A}{2} ) }{cos \frac{A}{2} + cos ^2 \frac{A}{2} +cos ^2 \frac{A}{2} }\\\\[/tex]
[tex]=\frac{sin\frac{A}{2}( 1 +2cos \frac{A}{2} ) }{cos \frac{A}{2} + 2cos ^2 \frac{A}{2} }\\\\=\frac{sin\frac{A}{2}( 1 +2cos \frac{A}{2} ) }{cos \frac{A}{2} ( 1 + 2cos \frac{A}{2}) }\\\\=\frac{sin\frac{A}{2}}{cos\frac{A}{2}}\\\\= tan \frac{A}{2}\\\\= RHS[/tex]