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

Sami Ullah

Last Activity: 6 Years ago

$First$ $cos\theta =2cos^{2}\frac{\theta}{2}-1$

$2cos^{2}\frac{\theta}{2}-1=\frac{cos\alpha cos\beta }{1-sin\alpha sin\beta }$

$2cos^{2}\frac{\theta}{2}=\frac{cos\alpha cos\beta }{1-sin\alpha sin\beta }+1$

$2cos^{2}\frac{\theta}{2}=\frac{cos\alpha cos\beta-sin\alpha sin\beta+1 }{1-sin\alpha sin\beta }$

$cos^{2}\frac{\theta}{2}=\frac{cos(\alpha +\beta )+1 }{2+(-2sin\alpha sin\beta )}$

$cos^{2}\frac{\theta}{2}=\frac{1+cos(\alpha +\beta )}{2+cos(\alpha+\beta )-cos(\alpha -\beta )}$

$sec^{2}\frac{\theta}{2}=\frac{2+cos(\alpha+\beta )-cos(\alpha -\beta)}{ 1+cos(\alpha +\beta)}$

$sec^{2}\frac{\theta}{2}-1=\frac{2+cos(\alpha+\beta )-cos(\alpha -\beta)}{ 1+cos(\alpha +\beta)}-1$

$tan^{2}\frac{\theta}{2}=\frac{2+cos(\alpha+\beta )-cos(\alpha -\beta)-1-cos(\alpha +\beta)}{1+cos(\alpha +\beta) }$

$tan^{2}\frac{\theta}{2}=\frac{1-cos(\alpha -\beta)}{1+cos(\alpha +\beta) }$

$tan^{2}\frac{\theta}{2}=\frac{2sin^{2}(\frac{\alpha -\beta }{2})}{2cos^{2}(\frac{\alpha +\beta }{2}) }$

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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Trigonometry> Please provide the solution of question n...

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

Paridhi sharma , 6 Years ago

Sami Ullah

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Trigonometry> Please provide the solution of question n...

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

Paridhi sharma , 6 Years ago

Sami Ullah

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