Question

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

Answer 1

$$\begin{gathered} \mathrm{Given}:\sec \theta +\tan \theta +1=0 \\ \sec \theta +\tan \theta +1=0 \\ \Rightarrow \sec \theta +\tan \theta =-1 \\ \mathrm{Multiplying}\mathrm{and}\mathrm{dividing}\mathrm{LHS}\mathrm{by}\sec \theta -\tan \theta ,\mathrm{we}\mathrm{get} \\ \Rightarrow \left(\sec \theta +\tan \theta \right)\times \left(\frac{\sec \theta -\tan \theta }{\sec \theta -\tan \theta }\right)=-1 \\ \Rightarrow \left(\frac{{\sec }^2\theta -{\tan }^2\theta }{\sec \theta -\tan \theta }\right)=-1 \\ \Rightarrow \left(\frac{1+{\tan }^2\theta -{\tan }^2\theta }{\sec \theta -\tan \theta }\right)=-1\left(\because {\sec }^2\theta =1+{\tan }^2\theta \right) \\ \Rightarrow \left(\frac{1}{\sec \theta -\tan \theta }\right)=-1 \\ \Rightarrow \left(\sec \theta -\tan \theta \right)=-1 \\ \mathrm{Hence},\mathrm{the}\mathrm{correct}\mathrm{option}\mathrm{is}\left(\mathrm{b}\right). \end{gathered}$$