Welcome :: Homework Help and Answers :: Mathskey.com

Q&A
Unanswered
Categories
Users
Ask a Question

Welcome to Mathskey.com Question & Answers Community. Ask any math/science homework question and receive answers from other members of the community.

Q&A Categories
BASIC MATH (93)
PRE-ALGEBRA (441)
ALGEBRA 1 (2,220)
GEOMETRY (864)
ALGEBRA 2 (2,751)
TRIGONOMETRY (1,187)
PRECALCULUS (1,903)
CALCULUS (2,802)
STATISTICS (129)
PHYSICS (659)
CHEMISTRY (137)
MECHANICAL ENGINEERING (64)
ELECTRICAL ENGINEERING (207)
Winplot Tutorials (2)

Most popular tags

solving-equations system-of-equations functions math slope-intercept-form physics homework-help trigonometric-identities integration limits substitution-method calculus

13,459 questions

17,854 answers

1,446 comments

811,117 users

cosx + sinx + tanx = secx verify this

0 votes

using trig identities

trigonometric-identities

asked Nov 30, 2013 in TRIGONOMETRY by homeworkhelp Mentor

Share this question

Google+ Facebook Twitter LinkedIn Reddit

Please log in or register to add a comment.

Please log in or register to answer this question.

1 Answer

0 votes

The left hand side identity

= Cosx+Sinx+Tanx

= (Cosx+Sinx+Sinx)/Cosx

= (Cos^2x+SinxCosx+Sinx)/Cosx

Cosx+Sinx+Tanx is not equals to secx.

answered Jan 30, 2014 by david Expert

Please log in or register to add a comment.

Related questions

[cosx(tanx-secx)]/(1-cscx)=sinx Verify identity

asked Mar 1, 2014 in TRIGONOMETRY by johnkelly Apprentice

trigonometric-identities
solving-trigonometric-equations

Verify the identity. cosx+sinx tanx=secx

asked Nov 20, 2013 in TRIGONOMETRY by abstain12 Apprentice

trigonometric-identities

How do I cerify the identity tanx sinx + cosx = secx

asked Feb 18, 2014 in TRIGONOMETRY by payton Apprentice

trigonometric-identities

Verify the identity: (sinx + tanx + tan^3x)/tanx = cosx + sec^2x

asked Nov 20, 2013 in TRIGONOMETRY by homeworkhelp Mentor

trigonometric-identities

prove that (secx+tanx)(1-sinx/cosx)=1

asked Aug 27, 2014 in PRECALCULUS by anonymous

trigonometric-identities
precalc
trig

1+secx/tanx+sinx

asked Nov 30, 2013 in TRIGONOMETRY by mathgirl Apprentice

trigonometric-identities

secx+tanx=cosx

asked Mar 1, 2014 in TRIGONOMETRY by payton Apprentice

trigonometric-identities
solving-trigonometric-equations

Verify the proof: (secx+tanx)/(cscx+1)=tanx.

asked Feb 12, 2014 in TRIGONOMETRY by mathgirl Apprentice

trigonometric-identities

Simplify the expression secx/sinx - sinx/cosx

asked Mar 1, 2014 in TRIGONOMETRY by mathgirl Apprentice

trigonometric-identities
solving-trigonometric-equations

(cosx)/(1-sinx)=2(secx)

asked Mar 1, 2014 in TRIGONOMETRY by rockstar Apprentice

trigonometric-identities
solving-trigonometric-equations

Prove sinx/cscx + cosx/secx = 1

asked Feb 24, 2014 in TRIGONOMETRY by linda Scholar

trigonometric-identities
solving-trigonometric-equations

Tan (x+π/4)= (cosx + sinx)/(cosx-sinx) How do I prove this?

asked Nov 13, 2014 in TRIGONOMETRY by anonymous

trigonometric-identities

Tan x/2 = (1-cosx+sinx)/(1+cosx+sinx) prove this?

asked Feb 11, 2014 in TRIGONOMETRY by futai Scholar

trigonometric-identities

how to verify this identity (cosxcotx)/(1-sinx)-1=cscx?

asked Nov 20, 2013 in TRIGONOMETRY by dkinz Apprentice

trigonometric-identities

Find tanx+secx

asked Sep 20, 2018 in TRIGONOMETRY by anonymous

trigonometric-identities
trignometric-ratios

Prove ((cscx-cotx)(cscx+cotx))/secx=cosx

asked Feb 24, 2014 in TRIGONOMETRY by dkinz Apprentice

trigonometric-identities
solving-trigonometric-equations

cosx(secx-cosx)=sin^2x

asked Nov 30, 2013 in TRIGONOMETRY by payton Apprentice

trigonometric-identities

varify the identity: csc^2(X) - (CosX)(SecX) = Cot^2(X)

asked Nov 20, 2013 in TRIGONOMETRY by mathgirl Apprentice

trigonometric-identities

(TanX-1)(CosX-1)=0

asked Feb 18, 2014 in TRIGONOMETRY by skylar Apprentice

trigonometric-identities

Let X and Y are real numbers such that sinX/sinY = 3 and cosX/cosY = 1/2.

asked Jun 24, 2014 in TRIGONOMETRY by anonymous

trigonometric-identities

Copyright © 2024 MATHSKEY. All Rights Reserved. Sitemap|Terms Of Use|Disclaimer|Contact Us|FAQ

...