Question

(1 + tan θ + sec θ) (1 + cot θ – cosec θ) = ?
(a) –1
(b) 0
(c) 1
(d) 2

Answer 1

$$\begin{gathered} \left(1+\tan \theta +\sec \theta \right)\left(1+\cot \theta -\mathrm{cosec}\theta \right) \\ =1+\cot \theta -\mathrm{cosec}\theta +\tan \theta +\tan \theta \cot \theta -\tan \theta \mathrm{cosec}\theta +\sec \theta +\sec \theta \cot \theta -\sec \theta \mathrm{cosec}\theta \\ =1+\frac{\cos \theta }{\sin \theta }-\frac{1}{\sin \theta }+\frac{\sin \theta }{\cos \theta }+\tan \theta \times \frac{1}{\tan \theta }-\frac{\sin \theta }{\cos \theta }\times \frac{1}{\sin \theta }+\frac{1}{\cos \theta }+\frac{1}{\cos \theta }\times \frac{\cos \theta }{\sin \theta }-\frac{1}{\cos \theta \sin \theta } \end{gathered}$$
$$\begin{gathered} =1+\frac{\cos \theta }{\sin \theta }-\frac{1}{\sin \theta }+\frac{\sin \theta }{\cos \theta }+1-\frac{1}{\cos \theta }+\frac{1}{\cos \theta }+\frac{1}{\sin \theta }-\frac{1}{\cos \theta \sin \theta } \\ =2+\frac{\cos \theta }{\sin \theta }+\frac{\sin \theta }{\cos \theta }-\frac{1}{\cos \theta \sin \theta } \\ =2+\frac{{\cos }^2\theta +{\sin }^2\theta }{\cos \theta \sin \theta }-\frac{1}{\cos \theta \sin \theta } \end{gathered}$$
$$\begin{gathered} =2+\frac{1}{\cos \theta \sin \theta }-\frac{1}{\cos \theta \sin \theta }\left({\cos }^2\theta +{\sin }^2\theta =1\right) \\ =2 \end{gathered}$$

Hence, the correct answer is option (d).