Question

root over of sec theta-1÷sec theta+1 + root over of sec rheta+1÷sec theta-1

Answer 1

Answer:

=\frac{2}{sin\alpha }=2cosec\alpha=\frac{2sec \alpha }{tan\alpha }

Step-by-step explanation:

$=\sqrt{\frac{sec \alpha-1}{sec \alpha+1}  }  +\sqrt{\frac{sec \alpha+1}{sec \alpha-1} } \\=\sqrt{\frac{(sec \alpha-1)(sec \alpha-1)}{(sec \alpha+1)(sec \alpha-1)} }  +\sqrt{\frac{(sec \alpha+1)(sec \alpha+1)}{(sec \alpha-1)(sec \alpha+1)} } \\=\sqrt{\frac{(sec \alpha-1)^{2} }{sec^{2}\alpha -1  } } +\sqrt{\frac{(sec \alpha+1)^{2} }{sec^{2}\alpha -1  } }\\Now, according to identity,1+tan^{2}\alpha  =sec^{2}\alpha$  

$=\sqrt{\frac{(sec \alpha-1)^{2} }{tan^{2}  } }+\sqrt{\frac{(sec \alpha+1)^{2} }{tan^{2}  } }\\=\frac{sec\alpha-1 }{tan\alpha } +\frac{sec\alpha+1 }{tan\alpha } \\=\frac{sec\alpha-1 +sec\alpha+1 }{tan\alpha } \\=\frac{2sec \alpha }{tan\alpha }$

Or if you want it in sin and cos ratio then:

$=\frac{2sec \alpha }{tan\alpha }=\frac{2(\frac{1}{cos\alpha }) }{\frac{sin\alpha }{cos\alpha }  } \\=\frac{2}{sin\alpha } \\or\\=2cosec\alpha$