# trigonometry...........

Prove the following identity:-
(i) tan^3x/1+tan^2x + cot^3x/1+cot^2 = 1-2sin^x cos^x/sinx cosx

1. 👍 0
2. 👎 0
3. 👁 170
1. You obviously did not read my reply to that same question, since you did not change the typing using brackets.

I still cannot tell who is divided by who ....

http://www.jiskha.com/display.cgi?id=1307704739

1. 👍 0
2. 👎 0

## Similar Questions

1. ### Inverse trigonometry

Prove that- tan^-1(1/2tan 2A)+tan^-1(cotA)+tan^-1(cot^3A) ={0,ifpi/4

2. ### Precalculus

Circle O below has radius 1. Eight segment lengths are labeled with lowercase letters. Six of these equal a trigonometric function of theta. Your answer to this problem should be a six letter sequence whose letters represent the

3. ### Trigonometry

Verify the identity algebraically. TAN X + COT Y/TAN X COT Y= TAN Y + COT X

4. ### Trigonometry

Equation 1: (tan^2x)/(cos^2x)+sec^2x+csc^2x=sec^4x+csc^2x*cos^2x+1 Equation 2: (csc^2x*(sin^4x+cos^2x))/(cos^2x)-cos^2x = tan^2x*csc^2x+cot^2x+sin^2x-1 Equation 3: tan^2x*sin^2x - (cot^2x)/(csc^2x) = -(sin^2x*cot^2x)/(csc^2x)-

1. ### Pre-calculus help

I have two problems I am stuck on, if you could show me how to solve the problems it would be much appreciated. 1) Find sin 2x, cos 2x, and tan 2x from the given information. tan x = − 1/6, cos x > 0 sin 2x = cos 2x = tan 2x =

2. ### Trig.......

I need to prove that the following is true. Thanks (2tanx /1-tan^x)+(1/2cos^2x-1)= (cosx+sinx)/(cosx - sinx) and thanks ........... check your typing. I tried 30º, the two sides are not equal, they differ by 1 oh , thank you Mr

3. ### Trigonometry

How do you verify the equation is an identity? Tan^2x-tan^2y=sec^2x-sec^2y and, how do you factor and simplify, cscx(sin^2x+cos^2xtanx)/sinx+cosx

4. ### 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

1. ### Trig

Use the fundamental identities to simplify the expression: cot beta sec beta I used 1+tan^2u=secu since cot is the inverse of tan. I flipped the tangent, then so it was 1+ (1/tan). But the book's answer is the cosecant of beta.

2. ### trig

For each expression in column I, choose the expression from column II to complete an identity: Column I Column II 1. -tanxcosx A. sin^2x/cos^2x 2. sec^2x-1 B. 1/sec^2x 3. sec x/cscx C. sin(-x) 4. 1+sin^2x D.csc^2x-cot^2x+sin^2x 5.

3. ### 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(x-y) = cos2y - cos2x My

4. ### math

prove (tan^3x/1 tan^2x) (cot^3x/1 cot^2) = (1-2sin^2x cos^2x)/sinx cosx