Question

tan theta+sectheta_1÷tantheta_sec theta+1=1+sintheta÷costheta

Answer 1

Here $LHS=\frac{\tan \theta +\sec \theta -1}{\tan \theta -\sec \theta +1} \\ LHS=\frac{\tan \theta +\sec \theta -1}{\tan \theta -(\sec \theta -1)} \\ LHS=\frac{(\tan \theta +\sec \theta -1)(\tan \theta +\sec \theta -1)}{(\tan \theta -(\sec \theta -1))(\tan \theta +\sec \theta -1)} \\ LHS=\frac{\tan^2 \theta +sec^2 \theta-2 \sec \theta +1+2\tan \theta (\sec \theta -1)}{\tan^2 \theta -\sec^2 \theta +2\sec \theta -1}$

$LHS=\frac{2\sec^2 \theta-2 \sec \theta +2\tan \theta (\sec \theta -1)}{2\sec \theta -2} \\ LHS=\frac{2( \sec \theta +\tan \theta) (\sec \theta -1)}{2(\sec \theta -1)} \\ LHS=\sec \theta +\tan \theta\\ LHS=\frac{1+\sin \theta}{\cos \theta}=RHS$

The proof is complete.