Question

Prove that \(\displaystyle \frac{\sin\theta-\cos\theta+1}{\sin\theta+\cos\theta-1}=\sec\theta+\tan\theta\).

Hint:

When RHS contains $\sec\theta+\tan\theta$, try rewriting it as $\dfrac{1+\sin\theta}{\cos\theta}$ or use half–angle substitutions $s=\sin\frac{\theta}{2},\,c=\cos\frac{\theta}{2}$ to simplify the LHS.

Answer 1

Using half–angle identities: $1-\cos\theta=2\sin^2\frac{\theta}{2}$ and $\sin\theta=2\sin\frac{\theta}{2}\cos\frac{\theta}{2}$. Put $s=\sin\frac{\theta}{2}$ and $c=\cos\frac{\theta}{2}$. Then
\[ \frac{\sin\theta-\cos\theta+1}{\sin\theta+\cos\theta-1} =\frac{2sc-(c^2-s^2)+1}{2sc+(c^2-s^2)-1} =\frac{2sc+2s^2}{2sc-2s^2} =\frac{c+s}{\,c-s\,}. \tag{1} \] Now, \[ \sec\theta+\tan\theta=\frac{1+\sin\theta}{\cos\theta} =\frac{1+2sc}{c^2-s^2} =\frac{(s+c)^2}{(c+s)(c-s)} =\frac{c+s}{\,c-s\,}. \tag{2} \] From (1) and (2), both sides are equal; hence proved.
\[ \boxed{\displaystyle \frac{\sin\theta-\cos\theta+1}{\sin\theta+\cos\theta-1}=\sec\theta+\tan\theta} \]