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Please provide the solution of question number 29 as soon as possible!

Please provide the solution of question number 29 as soon as possible! 

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Grade:11

1 Answers

Sami Ullah
46 Points
5 years ago
First        cos\theta =2cos^{2}\frac{\theta}{2}-1
 
So
 
2cos^{2}\frac{\theta}{2}-1=\frac{cos\alpha cos\beta }{1-sin\alpha sin\beta }
2cos^{2}\frac{\theta}{2}=\frac{cos\alpha cos\beta }{1-sin\alpha sin\beta }+1
2cos^{2}\frac{\theta}{2}=\frac{cos\alpha cos\beta-sin\alpha sin\beta+1 }{1-sin\alpha sin\beta }
cos^{2}\frac{\theta}{2}=\frac{cos(\alpha +\beta )+1 }{2+(-2sin\alpha sin\beta )}
cos^{2}\frac{\theta}{2}=\frac{1+cos(\alpha +\beta )}{2+cos(\alpha+\beta )-cos(\alpha -\beta )}
sec^{2}\frac{\theta}{2}=\frac{2+cos(\alpha+\beta )-cos(\alpha -\beta)}{ 1+cos(\alpha +\beta)}
sec^{2}\frac{\theta}{2}-1=\frac{2+cos(\alpha+\beta )-cos(\alpha -\beta)}{ 1+cos(\alpha +\beta)}-1
tan^{2}\frac{\theta}{2}=\frac{2+cos(\alpha+\beta )-cos(\alpha -\beta)-1-cos(\alpha +\beta)}{1+cos(\alpha +\beta) }
tan^{2}\frac{\theta}{2}=\frac{1-cos(\alpha -\beta)}{1+cos(\alpha +\beta) }
tan^{2}\frac{\theta}{2}=\frac{2sin^{2}(\frac{\alpha -\beta }{2})}{2cos^{2}(\frac{\alpha +\beta }{2}) }
Taking square root on both sides.
 
tan\frac{\theta}{2}=\frac{sin(\frac{\alpha -\beta }{2})}{cos(\frac{\alpha +\beta }{2}) }
tan\frac{\theta}{2}=\frac{sin(\frac{\alpha }{2}-\frac{\beta }{2})}{cos(\frac{\alpha }{2}+\frac{\beta }{2}) }
tan\frac{\theta}{2}=\frac{sin\frac{\alpha }{2}cos\frac{\beta }{2}-cos\frac{\alpha }{2}sin\frac{\beta }{2}}{cos\frac{\alpha }{2}cos\frac{\beta }{2}-sin\frac{\alpha }{2}sin\frac{\beta }{2} }
 
Dividing up and down by   cos\frac{\alpha }{2}cos\frac{\beta }{2},So we get ,
 
 
tan\frac{\theta}{2}=\frac{\frac{sin\frac{\alpha }{2}}{cos\frac{\alpha }{2}}-\frac{sin\frac{\beta }{2}}{cos\frac{\beta }{2}}}{1-\frac{sin\frac{\alpha }{2}sin\frac{\beta }{2}}{cos\frac{\alpha }{2}cos\frac{\beta }{2}} }
tan\frac{\theta}{2}=\frac{tan\frac{\alpha }{2}-tan\frac{\beta }{2}}{1-tan\frac{\alpha }{2}tan\frac{\beta }{2}}
 
And you are there.
This was what I could read in the question you posted.I didn’t get what was the whole question about.
Please correct my mistakes (IF ANY).  

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