Question

${\displaystyle \frac{\sin \theta }{(\sec \theta +\tan \theta -1)}+\frac{\cos \theta }{(\cos ec\theta +\cot \theta -1)}=1}$

Answer 1

LHS = ${\displaystyle \frac{\sin \theta }{(\sec \theta +\tan \theta -1)}+\frac{\cos \theta }{(\cos ec\theta +\cot \theta -1)}}$

       =${\displaystyle \frac{\sin \theta \cos \theta }{1+\sin \theta -\cos \theta }+\frac{\cos \theta \sin \theta }{1+\cos \theta -\sin \theta }}$

      =${\displaystyle \sin \theta \cos \theta [\frac{1}{1+(\sin \theta -\cos \theta )}+\frac{1}{1-(\sin \theta -\cos \theta )}]}$

      =${\displaystyle \sin \theta \cos \theta \left[\frac{1-(\sin \theta -\cos \theta )+1+(\sin \theta -\cos \theta )}{\{1+(\sin \theta -\cos \theta )\}\{1-(\sin \theta -\cos \theta )\}}\right]}$

     =${\displaystyle \sin \theta \cos \theta \left[\frac{1-\sin \theta +\cos \theta +1+\sin \theta -\cos \theta }{1-{(\sin \theta -\cos \theta )}^2}\right]}$

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

    =${\displaystyle \frac{2\sin \theta \cos \theta }{2\sin \theta \cos \theta }}$

    =1

    = RHS
Hence, LHS = RHS