Question

If $a=\sec \mathrm{\theta }-\tan \mathrm{\theta }\mathrm{and}b=\mathrm{cosec}\mathrm{\theta }+\cot \mathrm{\theta }$, then shown that $ab+a-b+1=0$.

Answer 1

$$\begin{gathered} a=sec\theta -\tan \theta \mathrm{And},b=\cos ec\theta +cot\theta \\ =\frac{1-\sin \theta }{\cos \theta }\mathrm{And},b=\frac{1+\cos \theta }{\sin \theta } \\ \mathrm{Now},\mathrm{we}\mathrm{have}: \\ ab+a-b+1 \\ \left(\frac{1-\sin \theta }{\cos \theta }\right)\left(\frac{1+\cos \theta }{\sin \theta }\right)+\frac{1-\sin \theta }{\cos \theta }-\left(\frac{1+\cos \theta }{\sin \theta }\right)+1 \\ =\frac{1-\sin \theta +\cos \theta -\sin \theta \cos \theta +\sin \theta -{\sin }^2\theta -\cos \theta -{\cos }^2\theta +\sin \theta \cos \theta }{\sin \theta \cos \theta } \\ =\frac{1-{\sin }^2\theta -{\cos }^2\theta }{\sin \theta \cos \theta } \\ =0 \end{gathered}$$

Hence proved.