Question

tanθ/secθ-1+tanθ/secθ+1 is equal to
A. 2 tan θ
B. 2 sec θ
C. 2 cosec θ
D. 2 tan θ sec θ

Answer 1

Tanθ/secθ-1+tanθ/secθ+1 is equal to

C. 2 cosec θ

• To solve this problem, we firstly need to know the trigonometric relationship between tan and sec i.e sec² x - tan² x = 1.   -(1)
• By this relation we can solve the given equation and equivalent relation can be.
• $\frac{tan x}{sec x -1} +\frac{tan x}{sec x +1}$
• $\frac{tan x (sec x -1 + sec x + 1)}{sec^{2} x -1}$
• From equation 1 -
• $\frac{tan x (2secx)}{tan^{2} x} = 2cosec x$

Answer 2

Step-by-step explanation:

Tanθ/secθ-1+tanθ/secθ+1 is equal to

C. 2 cosec θ

To solve this problem, we firstly need to know the trigonometric relationship between tan and sec i.e sec² x - tan² x = 1. -(1)

By this relation we can solve the given equation and equivalent relation can be.

\frac{tan x}{sec x -1} +\frac{tan x}{sec x +1}

secx−1

tanx

+

secx+1

tanx

\frac{tan x (sec x -1 + sec x + 1)}{sec^{2} x -1}

sec

2

x−1

tanx(secx−1+secx+1)

From equation 1 -

\frac{tan x (2secx)}{tan^{2} x} = 2cosec x

tan

2

x

tanx(2secx)

=2cosecθ (C)