Trig.......
 👍
 👎
 👁
Respond to this Question
Similar Questions

trigonometry
Prove: tan2x+sec2x = cosx+sinx/ cosxsinx

Trig
prove(sinx/cosx+1)+(cosx1/sinx)=0 Please help asap!!!!! PLEASE

Math
How do I solve this? tan^2x= 2tanxsinx My work so far: tan^2x  2tanxsinx=0 tanx(tanx  2sinx)=0 Then the solutions are: TanX=0 and sinX/cosX = 2 sin X Divide through by sinX: we have to check this later to see if allowed (ie sinX

Precalculus/Trig
I can't seem to prove these trig identities and would really appreciate help: 1. cosx + 1/sin^3x = cscx/1  cosx I changed the 1: cosx/sin^3x + sin^3x/sin^3x = cscx/1cosx Simplified: cosx + sin^3x/sin^3x = cscx/1cosx I don't

Calculus
Integrate sinx(cosx)^2dx using the substitution u=sinx. I know how to do this using u =cosx, but not sinx. The next problem on the homework was the same question except it asked to use u=cosx, so there couldn't have been a

Math
(sinx  cosx)(sinx + cosx) = 2sin^2x 1 I need some tips on trigonometric identities. Why shouldn't I just turn (sinx + cosx) into 1 and would it still have the same identity?

Math  Trig  Double Angles
Prove: sin2x / 1  cos2x = cotx My Attempt: LS: = 2sinxcosx /  1  (1  2sin^2x) = 2sinxcosx /  1 + 2sin^2x = cosx / sinx  1 = cosx / sinx  1/1 = cosx / sinx  sinx / sinx  Prove: 2sin(x+y)sin(xy) = cos2y  cos2x My

Math help again
cos(3π/4+x) + sin (3π/4 x) = 0 = cos(3π/4)cosx + sin(3π/4)sinx + sin(3π/4)cosx  cos(3π/4)sinx = 1/sqrt2cosx + 1/sqrt2sinx + 1/sqrt2cosx  (1/sqrt2sinx) I canceled out 1/sqrt2cosx and 1/sqrt2cosx Now I have 1/sqrt sinx +

Calculus
If y=3/(sinx+cosx) , find dy/dx A. 3sinx3cosx B. 3/(sinx+cosx)^2 C. 3/(sinx+cosx)^2 D. 3(cosxsinx)/(sinx+cosx)^2 E. 3(sinxcosx)/(1+2sinxcosx)

TRIGONOMETRY *(MATHS)
Q.1 Prove the following identities: (i) tan^3x/1+tan^2x + cot^3x/1+cot^2 = 12sin^x cos^x/sinx cosx (ii) (1+cotx+tanx)(sinxcosx)/sec^3xcosec^3x = sin^2xcos^2x.

trig
How do I prove these trig identities? secxcosx/tanx =sinx And 1+sinx/cosx+ cox/ 1+sinx=2secx

Maths
Solve this equation fo rx in the interval 0
You can view more similar questions or ask a new question.