Math, asked by mandalorian1905, 1 month ago

If cosθ = (2cosA-1)/(2-cosA), then tanθ/2 equals

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Answers

Answered by senboni123456
2

Step-by-step explanation:

We have,

 \tt{ \blue{cos( \theta) =  \dfrac{2 \: cos( \alpha ) - 1}{2 -  cos( \alpha ) }}}

 \tt{ \implies  \dfrac{1 -  tan^{2} \left( \dfrac{ \theta}{2}  \right) }{1  +  tan^{2} \left( \dfrac{ \theta}{2}  \right)} =  \dfrac{2  \left \{\dfrac{1 -  tan^{2} \left( \dfrac{  \alpha }{2}  \right) }{1  +  tan^{2} \left( \dfrac{ \alpha  }{2}  \right)} \right \} - 1}{2 -  \left \{\dfrac{1 -  tan^{2} \left( \dfrac{  \alpha }{2}  \right) }{1  +  tan^{2} \left( \dfrac{ \alpha  }{2}  \right)} \right \} }}

 \tt{ \implies  \dfrac{1 -  tan^{2} \left( \dfrac{ \theta}{2}  \right) }{1  +  tan^{2} \left( \dfrac{ \theta}{2}  \right)} =  \dfrac{2   \left \{1 -  tan^{2} \left( \dfrac{  \alpha }{2}  \right) \right \} -\left \{1  +  tan^{2} \left( \dfrac{  \alpha }{2}  \right) \right \}  }{2   \left \{1  +   tan^{2} \left( \dfrac{  \alpha }{2}  \right) \right \} -\left \{1   -   tan^{2} \left( \dfrac{  \alpha }{2}  \right) \right \} }}

 \tt{ \implies  \dfrac{1 -  tan^{2} \left( \dfrac{ \theta}{2}  \right) }{1  +  tan^{2} \left( \dfrac{ \theta}{2}  \right)} =  \dfrac{2  - 2 \:  tan^{2} \left( \dfrac{  \alpha }{2}  \right)  -1   -  tan^{2} \left( \dfrac{  \alpha }{2}  \right)   }{2    +  2 \:  tan^{2} \left( \dfrac{  \alpha }{2}  \right)  -1    +   tan^{2} \left( \dfrac{  \alpha }{2}  \right)  }}

 \tt{ \implies  \dfrac{1 -  tan^{2} \left( \dfrac{ \theta}{2}  \right) }{1  +  tan^{2} \left( \dfrac{ \theta}{2}  \right)} =  \dfrac{1  - 3 \:  tan^{2} \left( \dfrac{  \alpha }{2}  \right)       }{1    +  3 \:  tan^{2} \left( \dfrac{  \alpha }{2}  \right)   }}

Using componendo and dividendo,

 \tt{ \implies  \dfrac{1 -  tan^{2} \left( \dfrac{ \theta}{2}  \right) + 1  +  tan^{2} \left( \dfrac{ \theta}{2}  \right) }{1 -  tan^{2} \left( \dfrac{ \theta}{2}  \right)  -  1   -   tan^{2} \left( \dfrac{ \theta}{2}  \right)} =  \dfrac{1  - 3 \:  tan^{2} \left( \dfrac{  \alpha }{2}  \right)   +   1    +  3 \:  tan^{2} \left( \dfrac{  \alpha }{2}  \right)     }{1  - 3 \:  tan^{2} \left( \dfrac{  \alpha }{2}  \right)    -   1     -   3 \:  tan^{2} \left( \dfrac{  \alpha }{2}  \right)    }}

 \tt{ \implies  \dfrac{2  }{ -  2tan^{2} \left( \dfrac{ \theta}{2}  \right)    } =  \dfrac{2  }{ - 6\:  tan^{2} \left( \dfrac{  \alpha }{2}  \right)     }}

 \tt{ \implies  \dfrac{1  }{ tan^{2} \left( \dfrac{ \theta}{2}  \right)    } =  \dfrac{1  }{ 3\:  tan^{2} \left( \dfrac{  \alpha }{2}  \right)     }}

 \tt{ \implies  tan \left( \dfrac{ \theta}{2}  \right)     =  \pm \sqrt{3}\:  tan \left( \dfrac{  \alpha }{2}  \right)     }

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