The domain of the function f x - x -tan x isA- R -2 n -1 -2- n - Z B- R - n - n - Z C- R -0D- R

The correct option is A ℝ {(2n+1)π2, n∈ℤ}Domain of nbsp;x=ℝ {(2n+1)π2, n∈ℤ}=D_1 Domain of nbsp;tan x=ℝ {(2n+1)π2, n∈ℤ}=D_2 ∴ Domain of nbsp;f(x)=D_1∩ D_2 ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Mathematics

Domain and Range of Trigonometric Ratios

The domain of...

Question

The domain of the function $f (x) = sec x + tan x$ is

R - {(2 n + 1) \frac{π}{2}, n \in Z}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

R - {n π, n \in Z}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

R - {0}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

R

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is A $R - {(2 n + 1) \frac{π}{2}, n \in Z}$
Domain of $sec x = R - {(2 n + 1) \frac{π}{2}, n \in Z} = D_{1}$
Domain of $tan x = R - {(2 n + 1) \frac{π}{2}, n \in Z} = D_{2}$
$∴$ Domain of $f (x) = D_{1} \cap D_{2} = R - {(2 n + 1) \frac{π}{2}, n \in Z}$

Suggest Corrections

Similar questions

Q. The function f (x) = |cos x| is
(a) differentiable at x = (2n + 1) π/2, n ∈ Z
(b) continuous but not differentiable at x = (2n + 1) π/2, n ∈ Z
(c) neither differentiable nor continuous at x = n ∈ Z
(d) none of these

Q. If

| z_{1} + z_{2} | = | z_{1} - z_{2} |

then

arg z_{1} - arg z_{2} =

Q. The general solution(s) of

{cos}^{2} 2 x - {sin}^{2} x = 0

can be

Q. The function f(x) = cot x is discontinuous on the set

(a) {x : x = nπ, n ∈ Z}

(b) {x : x = 2nπ, n ∈ Z}

(C)

\{x : x = (2 n + 1) \frac{π}{2}, n \in Z\}

(d)

\{x : x = \frac{n π}{2}, n \in Z\}

Q. If

f (x) = \frac{7}{x} + \frac{2}{x + 1} + \frac{1}{x - 1},

then domain of the function is

Join BYJU'S Learning Program

Explore more

Domain and Range of Trigonometric Ratios

Standard XII Mathematics

Join BYJU'S Learning Program

The domain of the function f(x)=secx+tanx is

The correct option is A R−{(2n+1)π2, n∈Z}Domain of secx=R−{(2n+1)π2, n∈Z}=D1 Domain of tanx=R−{(2n+1)π2, n∈Z}=D2 ∴ Domain of f(x)=D1∩D2=R−{(2n+1)π2, n∈Z}

The domain of the function $f (x) = sec x + tan x$ is

The correct option is A $R - {(2 n + 1) \frac{π}{2}, n \in Z}$
Domain of $sec x = R - {(2 n + 1) \frac{π}{2}, n \in Z} = D_{1}$
Domain of $tan x = R - {(2 n + 1) \frac{π}{2}, n \in Z} = D_{2}$
$∴$ Domain of $f (x) = D_{1} \cap D_{2} = R - {(2 n + 1) \frac{π}{2}, n \in Z}$