If tan θ + sec θ = ex , then cos θ equal is

Question

If tan θ + sec θ = ex , then cos θ equal is (A) (ex + e-x/2) (B) (2/ex + e-x) (C) (ex - e-x/2) (D) (ex - e-x/ex + e-x)

Tardigrade Admin · Accepted Answer

Here, sec θ + tan θ = ex ....(i) Also sec2 θ - tan2 θ = 1 ....(ii) Dividing (it) by (i), we get sec θ - tan θ = e-x ...(iii) Add (i) and (iii) to obtain sec θ = (ex + e-x/2)

Q. If $tan θ + sec θ = e^{x}$ , then $cos θ$ equal is

$\frac{e ^{x} + e ^{- x}}{2}$

$\frac{2}{e ^{x} + e ^{- x}}$

$\frac{e ^{x} - e ^{- x}}{2}$

$\frac{e ^{x} - e ^{- x}}{e ^{x} + e ^{- x}}$

Here, $sec θ + tan θ = e^{x}$ ....(i)
Also $sec^{2} θ - tan^{2} θ = 1$ ....(ii)
Dividing (it) by (i), we get
$sec θ - tan θ = e^{- x}$ ...(iii)
Add (i) and (iii) to obtain $sec θ = \frac{e ^{x} + e ^{- x}}{2}$

Q. If tanθ+secθ=ex , then cosθ equal is

2ex+e−x​

ex+e−x2​

2ex−e−x​

ex+e−xex−e−x​