BD Merit - Freelancing, BCS, Bank, Primary, Admission, NTRCA

Discuss Forum

1.

A. $\sec^{2} θ = 1 + \tan^{2} θ$
B. $\sec^{2} θ = 1 + \tan^{2} θ$
C. $\sec^{2} θ = 1 + \tan^{2} θ$
D. $\sec^{2} θ = 1 + \tan^{2} θ$

Answer: Option A

Explanation:

আমরা জানি,

sec2o - tan2o = 1

so, sec2o = 1 + tan2o

Post your comments here:

Name:

Mobile:

Email:(Optional)

» Your comments will be displayed only after manual approval.