Log (cosβ) = p ⇒ cos β = 10^p ∴ secβ = (1/(cosβ)) = (1/(10^p )) = 10^(−p) ∴ Log (secβ) = Log 10^(−p) = −p Log 10 = −p


Question and Answers Forum

All Questions Topic List
Algebra Questions
Previous in All Question Next in All Question
Previous in Algebra Next in Algebra
Question Number 117781 by hmh2k20 last updated on 13/Oct/20

$Log (\cos β) = p \Rightarrow \cos β = 10^{p}$ $∴ \sec β = \frac{1}{\cos β} = \frac{1}{10^{p}} = 10^{- p}$ $∴ Log (\sec β) = Log 10^{- p}$ $= - p Log 10$ $= - p$
	Answered by $@y@m last updated on 13/Oct/20

	$K i n d l y u s e e d i t o r o p t i o n o f t h e a p p$ $f o r r o u g h w o r k .$ $J u s t t a p o n \equiv s i g n o n t o p l e f t s i d e$ $a n d t h e n t a p o n editor .$ $T h i s s e c t i o n i s m e a n t f o r m u t u a l$ $d i s c u s s i o n .$
	Commented by Tinku Tara last updated on 13/Oct/20

	$Latest updated app opens forum$ $be default and editor is opened$ $from navigation window .$ $Most of users are using the app$ $only to navigate forum and a$ $few actually create user id . This$ $interface simplfies browsing for$ $majority of users .$ $Only a small set of users are actually$ $using offline editor .$
	Commented by $@y@m last updated on 13/Oct/20

	$Y o u a r e r i g h t .$ $O f f l i n e e d i t o r i s u s e d r a r e l y .$ $S o I a p p r e c i a t e n e w u p d a t e .$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum