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

Question:

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

Solution:

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).

AYAN
Dec. 9, 2023, 6:35 a.m.
1 + tan²θ = sec² 1 = sec²θ - tan²θ sec²θ - tan²θ = ( secθ + tanθ)(secθ - tanθ)..........( Using a² - b² Identity) Also Secθ + tanθ + 1 = 0 Therefore, secθ + tanθ = - 1 Now sec²θ - tan²θ = ( secθ + tanθ)(secθ - tanθ) 1 = ( - 1)( secθ - tanθ) 1/(-1) = secθ - tanθ => Secθ - tanθ = -1 Hence proved 🫡 Correct option.............>(B)
AYAN
Dec. 9, 2023, 6:35 a.m.
1 + tan²θ = sec² 1 = sec²θ - tan²θ sec²θ - tan²θ = ( secθ + tanθ)(secθ - tanθ)..........( Using a² - b² Identity) Also Secθ + tanθ + 1 = 0 Therefore, secθ + tanθ = - 1 Now sec²θ - tan²θ = ( secθ + tanθ)(secθ - tanθ) 1 = ( - 1)( secθ - tanθ) 1/(-1) = secθ - tanθ => Secθ - tanθ = -1 Hence proved 🫡 Correct option.............>(B) ???? What about this explanation of mine Brother
AYAN
Dec. 9, 2023, 6:35 a.m.
1 + tan²θ = sec² 1 = sec²θ - tan²θ sec²θ - tan²θ = ( secθ + tanθ)(secθ - tanθ)..........( Using a² - b² Identity) Also Secθ + tanθ + 1 = 0 Therefore, secθ + tanθ = - 1 Now sec²θ - tan²θ = ( secθ + tanθ)(secθ - tanθ) 1 = ( - 1)( secθ - tanθ) 1/(-1) = secθ - tanθ => Secθ - tanθ = -1 Hence proved 🫡 Correct option.............>(B)
Narayani
March 20, 2023, 6:35 a.m.
Tera baap
Oct. 25, 2023, 6:35 a.m.
Hatt
AYAN
Dec. 9, 2023, 6:35 a.m.
1 + tan²θ = sec² 1 = sec²θ - tan²θ sec²θ - tan²θ = ( secθ + tanθ)(secθ - tanθ)..........( Using a² - b² Identity) Also Secθ + tanθ + 1 = 0 Therefore, secθ + tanθ = - 1 Now sec²θ - tan²θ = ( secθ + tanθ)(secθ - tanθ) 1 = ( - 1)( secθ - tanθ) 1/(-1) = secθ - tanθ => Secθ - tanθ = -1 Hence proved 🫡 Correct option.............>(B) ???? What about this explanation of mine Brother
Jan
March 16, 2023, 6:35 a.m.
🥰🥰🥰
Jan
March 16, 2023, 6:35 a.m.
🥰🥰🥰