If sec θ + tan θ + 1 = 0 then (sec θ – tan θ) = ?

Given: $\sec \theta+\tan \theta+1=0$

$\sec \theta+\tan \theta+1=0$

$\Rightarrow \sec \theta+\tan \theta=-1$

Multiplying and dividing LHS by $\sec \theta-\tan \theta$, we get

$\Rightarrow(\sec \theta+\tan \theta) \times\left(\frac{\sec \theta-\tan \theta}{\sec \theta-\tan \theta}\right)=-1$

$\Rightarrow\left(\frac{\sec ^{2} \theta-\tan ^{2} \theta}{\sec \theta-\tan \theta}\right)=-1$

$\Rightarrow\left(\frac{1+\tan ^{2} \theta-\tan ^{2} \theta}{\sec \theta-\tan \theta}\right)=-1 \quad\left(\because \sec ^{2} \theta=1+\tan ^{2} \theta\right)$

$\Rightarrow\left(\frac{1}{\sec \theta-\tan \theta}\right)=-1$

$\Rightarrow(\sec \theta-\tan \theta)=-1$

Hence, the correct option is (b).