Wednesday

April 23, 2014

April 23, 2014

Number of results: 4,731

**Trig**

draw the triangles and you will see that tan(s) = 12/5 tan(t) = 4/3 Now just plug in your sum formula: tan(s+t) = (tan s + tan t)/(1 - tan s * tan t)
*Tuesday, March 19, 2013 at 3:22pm by Steve*

**trig**

tan(x+y)=(tan(x)+tan(y))/(1-tan(x)tan(y)) substitute x=y to get tan(2x) = 2tan(x)/(1-tan²(x)) Substitute tan(2x)=1 and solve for tan(x). Reject the negative root knowing that tan(x)≥0 for 0<x<π/2.
*Tuesday, August 24, 2010 at 3:40am by MathMate*

**maths --plse help me..**

make a sketch, label the height as h distance from last position to the tower as x tan (alpha) = h/(x+a) h = (x+a)tan alpha in the same way h = xtan beta (x+a)tan alpha = xtan beta x tan alpha - xtan beta = - a tan alpha x(tan alpah - tan beta) = -a tan alpha x = -a tan alpha...
*Thursday, December 13, 2012 at 12:22pm by Reiny*

**Math**

tan (pi/4+x) = (tan pi/4 + tan x)/(1 -tan pi/4 tan x) tan (pi/4+x) = (tan pi/4 - tan x)/(1 +tan pi/4 tan x) note tan pi/4 = 1 add and put over common denominator [(1+tan x)(1+tan x)+(1-tan x)(1-tan x)]/(1 - tan^2 x) expand top [1+2 tanx +tan^2 x +1 - 2 tan x +tan^2 x]/(1 - tan...
*Sunday, January 4, 2009 at 2:48am by Damon*

**Trigonometry**

(1+tan)/(1-tan) multiply top and bottom by 1+tan (1+tan)^2 / (1-tan)(1+tan) (1 + 2tan + tan^2)/(1-tan^2) now, since sec^2 = 1+tan^2, (sec^2 + 2tan)/(1-tan^2)
*Friday, August 3, 2012 at 3:31pm by Steve*

**pre calc**

tan^5 - 9tan = 0 tan(tan^4-9) = 0 tan(tan^2-3)(tan^2+3) = 0 tan(tan-√3)(tan+√3)(tan^2+9) = 0 so, tan x = 0 or √3 or -√3 now it should be a cinch
*Sunday, April 14, 2013 at 5:02pm by Steve*

**Math-Trigonometry**

Show that if A, B, and C are the angles of an acute triangle, then tan A + tan B + tan C = tan A tan B tan C. I tried drawing perpendiculars and stuff but it doesn't seem to work? For me, the trig identities don't seem to plug in as well. Help is appreciated, thanks.
*Wednesday, November 6, 2013 at 5:21pm by Sam*

**Pre-cal**

since sec^2 x = 1+tan^2 x, we have 2(1+tan^2 x)+tan^2 x-3=0 3tan^2 x-1 = 0 tan^2 x = 1/3 tan x = ±1/√3 tan x > 0 in QI, QIII, so x = π/6, 7π/6 tan x < 0 in QII, QIV, so x = 5π/6, 11π/6 Actually, since the period of tan x is π, it would be ...
*Tuesday, July 2, 2013 at 5:10am by Steve*

**Precalculus II**

recall that sec^2 = 1+tan^2, so we have tan^2(x) + 3tan(x) - 4 = 0 (tan(x)+4)(tan(x)-1) = 0 tan(x) = 1 or -4 So, there will be 4 solutions, one in each quadrant.
*Wednesday, March 26, 2014 at 1:24pm by Steve*

**trig**

tan(x) has a period of π, so tan(x)=tan(x+π), in fact, tan(x+kπ)=tan(x), where k∈Z So to evaluate tan(257π/4) =tan((257/4)π) =tan(64π + π/4) =tan(π/4) I will let you figure out which answer to choose.
*Wednesday, May 23, 2012 at 9:44pm by MathMate*

**calculus**

can anyone tell me if tan-1x= 1 over tan x? No. They are different. tan^-1 (x) is a frequently used way of writing arctan x, which means the angle whose tangent is x. tan (tan^-1 x) = x
*Sunday, February 11, 2007 at 12:16pm by Elizabeth*

**Trig**

Did you recoginize that your expression matches the formula for tan 2A ? tan 2A = 2tanA/(1-tan^2 A) so 2tan(5pi/12)/1-tan²(5pi/12) = tan(5pi/6) or tan 150º tan 150 = -tan 30 or - tan(pi/6) = -1/√3
*Sunday, April 19, 2009 at 2:51pm by Reiny*

**Math**

tan^-1(1/2)=x=.463648 tan^-1(2/3)=y.588003 x+y+pi/2 = -z=2.62245 z=-2.62245 cos (z) = -.868243 and tan a= -1/2 tan b= -2/3 tan (a+b)=(tan a +tan b)/(1-tan a.tan b) so tan (a+b)= -7/4 sec^2 x-tan^2 x=1 so sec (a+b)=√(1+49/16) =√(65/16) (a+b)=1/sec (a+b) so cos(a+b) ...
*Tuesday, June 9, 2009 at 1:36pm by Sophie*

**trig**

recall tan(A+B) = (tanA + tanB)/(1 - tanAtanB) I believe you meant to type (tan 25° + tan 5°) /( 1- tan 25° tan 5°) now it fits the above pattern and it equals tan(25+5) = tan 30° from my 30-60-90 triangle I know that tan 30° = 1/√3 or √3/3
*Friday, April 30, 2010 at 10:07pm by Reiny*

**Pre-Calc (Prove the identity)**

Recall that formula for sum in tangent: tan (a +/- b) = [tan(a) +/- tan(b)] / [1 -/+ tan(a)tan(b)] Applying this, tan^2(x/2 + π/4) = { [tan(x/2) + tan(π/4)] / [1 - tan(x/2)tan(π/4)] }^2 Note that tan(π/4) = 1, so = [(tan(x/2) + 1) / (1 - tan(x/2))]^2 = (tan...
*Sunday, November 24, 2013 at 4:29pm by Jai*

**pre-calc**

1 + 2 tan x + tan^2 x = 2 tan x + 2 tan^2 x - 1 = 0 tan x = 1 or -1 x = 45 or 135 or or 225 or 315 degrees
*Tuesday, November 20, 2012 at 3:20pm by Damon*

**precal**

LS = tan(3x) = tan(2x + x) = (tan2x + tanx)/(1- tan2x tanx) = [2tanx/(1-tan^2 x) + tanx ] / [(1 - 2tanx(tanx)/(1 - tan^2 x) ] multiply top and bottom by 1 - tan^2 x (2tanx + tanx + tan^3 x)/(1 - tan^2 x - tan^2 x) = (3tanx + tan^3 x)/(1 - 3tan^2 x) = RS
*Monday, February 6, 2012 at 1:56pm by Reiny*

**Precalculus**

Hint: use the following trigonometric identity tan(a+b)=(tan(a)+tan(b))/(1-tan(a)tan(b))
*Wednesday, February 23, 2011 at 7:17pm by MathMate*

**Math**

Unfortunately it doesn't look right, ad... from the very start. Could you rechceck your source from where you got the expression? [ 1 - (tan α)(tan β) ] / [ 1 + (tan α)(tan β)] represents, according to me, cos(α+β)/cos(α-β). All is not ...
*Tuesday, June 9, 2009 at 1:36pm by MathMate*

**Trigonometry**

Find the exact value of tan(a-b) sin a = 4/5, -3pi/2<a<-pi; tan b = -sqrt2, pi/2<b<pi identity used is: tan(a-b)=(tan a-tan b)/1+tan a tan b simplify answer using radicals. (a is alpha, b is beta)
*Monday, November 8, 2010 at 11:40pm by Dee*

**Trig.**

sec^2xcotx-cotx=tanx (1/cos)^2 times (1/tan)-(1/tan)=tan (1/cos^2) times (-2/tan)=tan (-2/cos^2tan)times tan=tan(tan) sq. root of (-2/cos^2)= sq. root of (tan^2) sq. root of (2i)/cos=tan I'm not sure if I did this right. If I didn't, can you show me the correct steps? Thanks, ...
*Saturday, January 9, 2010 at 10:27am by Trig*

**math**

In most cases I suggest changing everything to sines and cosines, but sometimes the variations of the Pythagorean identities come in handy that is, by dividing sin^2 A + cos^2 A = 1 by cos^2 A we get tan^2 A + 1 = sec^2 A so the denominator of the first one is -tan^2 e and the...
*Sunday, February 22, 2009 at 12:41pm by Reiny*

**precalculus**

For each of the following determine whether or not it is an identity and prove your result. a. cos(x)sec(x)-sin^2(x)=cos^2(x) b. tan(x+(pi/4))= (tan(x)+1)/(1-tan(x)) c. (cos(x+y))/(cos(x-y))= (1-tan(x)tan(y))/(1+tan(x)tan(y)) d. (tan(x)+sin(x))/(1+cos(x))=tan(x) e. (sin(x-y...
*Sunday, April 14, 2013 at 3:27pm by anonymous*

**trig**

Both cot x and tan x must either be 1 or -1, since tan x = 1/cot x tan^2 x = 1 tan x = +/-1 tan 1 = 1 at 45 and 225 degrees. tan x = -1 at 135 and 315 degrees.
*Monday, February 18, 2008 at 11:05pm by drwls*

**Trig**

Using tan(A-B) = (tanA - tanB)/(1+tanAtanB) RS = tan(x-6π) = (tanx - tan 6π)/(1 + tanxtan 6π) but 6π is coterminal with 2π (6π is 3 rotations, and 2π is one rotation so tan 6π = tan 2π = 0 = (tanx - 0)/(1+ ) = tanx = LS so the ...
*Wednesday, January 23, 2013 at 8:14pm by Reiny*

**math**

if you need tan 7.5 degrees you would get .1316525 if you want tan 7.5 radians that would be 2.706 if the question asked for the "exact" value of tan 7.5º you would first find tan 15º by using tan 15º = tan(45º - 30º) = (tan45 - tan30)/(1 + tan45tan30) = (√3 - 1)?)√...
*Saturday, August 15, 2009 at 10:07pm by Reiny*

**calculus**

If limit as delta x approaches 0 of tan(0+Δx)-tan(0)/Δx =1 which of the following is false: d/dx [tanx]=1 the slope of y = tan(x) at x = 0 is 1 y = tan(x) is continuous at x = 0 y = tan(x) is differentiable at x = 0
*Thursday, February 6, 2014 at 12:56pm by lisa*

**trig**

h t t p : / / i m g 4 0 . i m a g e s h a c k . u s / c o n t e n t . p h p ? p a g e = d o n e& l = i m g 4 0 / 1 4 8 / a s d a s d y e . j p g & v i a = m u p l o a d (def of tan theta = a^-1 o)a = o opposite = adjacent tan theta written with respect to the first angle ...
*Sunday, September 6, 2009 at 5:06pm by trig*

**Grd 12 Math**

tan (pi-x) = -tan(x) cot(pi/2+x) = -tan(x) tan(2pi-x) = -tan(x) now things fall out.
*Wednesday, November 6, 2013 at 9:17am by Steve*

**Pre-Calculus**

your text should have tan(a-b) = (tan a - tan b)/(1 + tanatanb) so you will need tan a and tan b cos a = -3/5 and a is in the second quadrant, so by sketching a diagram it is easy to see that sin a = 4/5 and tan a = - 4/3 similarly sin b = 5/13 and b is in quadrant II, so cos ...
*Saturday, August 30, 2008 at 10:12pm by Reiny*

**calculus**

Actually, since tan(pi/2-x) = cot(x) and cot = 1/tan it falls right out. Or, using the sum of tangents formula, tan(arctanx + arctan(1/x)) = [tan(arctan(x)) + tan(arctan(1/x))][1 - tan(arctan(x))*tan(arctan(1/x))] = [x + 1/x]/[1 - x*1/x] = (x + 1/x)/0 = oo tan pi/2 = oo
*Wednesday, December 7, 2011 at 3:59pm by Anonymous*

**trig**

tan(345)? Tan(2*345)=tan(690)=tan(720-30)=tan(-30) tan(2*345)= 2*tan(345)/(1-tan^2(345) lets call tan(345) Y tan(-30)=2y/(1-y^2) Call Tan(-30) X x-XY^2=2Y XY^2+2Y-x=0 use the quadratic to solve for y in terms of X (tan(-30)=-1/sqrt3)
*Thursday, November 19, 2009 at 7:38pm by bobpursley*

**Pre-Calculus**

LS: sec/(sec-tan) multiply by (sec+tan) top and bottom sec(sec+tan) / (sec^2-tan^2) now recall that sec^2 = 1+tan^2, so we have (sec^2 + sec tan) / 1 sec^2 + sec tan = RS
*Monday, November 12, 2012 at 1:28pm by Steve*

**math**

Using reference angles, (take 225 and subtract 180), you get 45. Since 225° is in quadrant 3, it is positive. Now you have 1 + tan^2 (45°)… Tan^2 is the same as tan x tan, so you have 1 + tan(45) x tan (45). If you use a 45-45-90 triangle, you get tan(45) =1. So, 1 + (1x1). ...
*Sunday, September 23, 2012 at 9:05am by Monica*

**trig**

we know that tan 2A = 2tanA/(1 - tan^2 A) so tan 210 = 2tan105/(1 - tan^2 105) you should know that tan 210 = tan 30° = 1/√3 let tan 105 = x then 1/√3 = 2x/(1-x^2) 2√3x = 1 - x^2 x^2 + 2√3 - 1 = 0 x = (-2√3 ± √16)/2 = -√3 ± 2 but ...
*Monday, July 22, 2013 at 8:10pm by Reiny*

**calculus**

The angle joining P to the origin is A = tan^-1 (y/x) You can invert that to read y = x tan A In addition to that, along the parabola, y = x^2 Therefore x = tan A and y = tan^2 A
*Wednesday, September 8, 2010 at 10:14pm by drwls*

**math**

Volume = Integral pi [y(x)]^2 dx x = 0 to pi/4 The integral of tan^4(x), which you will need, can be found in a table of integrals. It uses a recursion formula. INT tan^4(x) = (1/3)tan^3x - (1/2)tan^2x + tan x - x
*Friday, January 30, 2009 at 9:13pm by drwls*

**Pre calc**

1. 2 x 67.5 = 135 2. tan 135 = - tan 45 = -1 3. tan 2A = [2tan A]/[1-(tan A)^2] 4. Put A = 67.5 in (3) and get tan135 = [2tan67.5]/[1-(tan67.5)^2] -1 = [2tan 67.5]/[1-(tan 67.5)^2] 5. Let x = tan 67.5. Then, -1 = 2x/[1 - x^2] x^2 - 2x - 1 = 0 By quadratic formula, x = 1 + sqrt...
*Sunday, January 22, 2012 at 10:42pm by Mick from cannotta*

**trig**

tangent has a period of π so tan(x) = tan(x+π) = tan(x + 2π) ... Also, tan(x)=-tan(π - x) so tan(23π/6) =tan(5π/6 + 3π) =tan(5π/6) =-tan(π/6) =-√3/3 A reference angle is the acute angle between the terminal side of the angle and the x-axis. So ...
*Sunday, October 4, 2009 at 8:17pm by MathMate*

**Trig-Weird Geometry Problem**

Determine all triangles ABC for which tan(A-B)+tan(B-C)+tan(C-A)=0. There's a hint: "Can you relate A-B to B-C and C-A?" Should I apply the tangent difference formula (tan(x-y))? Help would be appreciated, thanks.
*Thursday, December 5, 2013 at 11:02am by Sam*

**Pre-Calc**

tan(255) = tan(180+75) = tan 75 = tan(30 + 45) = (tan30+tan45)/ (1-tan30tan45) You can finish the rest. Note: tan(180+75) = (tan180+tan75)/ (1-tan180tan75) = (0+ tan75)/(1-0) =tan 75 = tan (30+45) not my answer but credits to: mlam18
*Wednesday, December 18, 2013 at 5:34pm by Danny Simone*

**Pre-Calculus**

So we must transform tan(A) + cot(A) into csc(A)*sec(A). tan(A) + cot(A) Note that cot(A) is equal to 1/tan(A). Substituting, = tan(A) + 1/tan(A) = (tan^2 (A) + 1) / tan(A) Recall the pythagorean identity: tan^2 (A) + 1 = sec^2 (A). Also tan(A) = sin(A)/cos(A) and sec(A) = 1/...
*Sunday, October 27, 2013 at 3:28pm by Jai*

** Math**

cot(4c-π/4) = 1/tan(4c-π/4) = 1/[ (tan4c) - tanπ/4)/(1 + tan4c(tanπ/4))] = (1+tan4ctanπ/4)/(tan4c - tanπ/4) = (1 + tan4c)/(tan4c - 1) , since tan π/4 = 1 also from tan 2A = 2tanA/(1 - tan^2 A) the above is = [ 1 + 2tan2c/(1 - tan^2 2c) ] / [...
*Saturday, June 18, 2011 at 10:50pm by Reiny*

**trig**

Tan7pi/4=Tan(pi+3pi/4) lies in IVQdt and is =-Tan pi/4. Tan 5pi/4=tan (pi+pi/4) lis in IIIQdt and is=tan pi/4, hence your ans is correct.
*Thursday, July 18, 2013 at 10:49pm by Anonymous*

**math**

because there is an exact relationship between π and the circle, thus its angles e.g. tan 45° = tan π/4 or tan .7854 ( I had to round it off) set your calculator to radians tan π.4 = 1 tan .7854 = 1.000003673
*Monday, March 25, 2013 at 9:08pm by Reiny*

**Integration**

Intergrate ¡́ sec^3(x) dx could anybody please check this answer. are the steps correct? thanks. = ¡́ sec x d tan x = sec x tan x - ¡́ tan x d sec x = sec x tan x - ¡́ sec x tan^2(x) dx = sec x tan x + ¡́ sec x dx - ¡́ sec^3(x) dx = sec x tan x + ln |sec x + tan x| - ¡́ sec^3(...
*Sunday, October 15, 2006 at 2:06am by Vidal*

**Precalculus troubles**

It will be the same as -tan 15. They probably want you to use the formula tan(x+y) = [tanx + tany]/[1 - tanx*tany] tan 300 = -tan 60 = -sqrt3 tan 45 = 1 tan 345 = (1 - sqrt3)/[1 + sqrt3) = (1 - sqrt3)^2/(1-3) = (-1/2)(sqrt3 -1)^2 = -0.26795
*Tuesday, January 12, 2010 at 11:15pm by drwls*

**Calculus**

use tan^2x = sec^2x - 1 tan^6 = tan^4 (sec^2 - 1) = tan^4 sec^2 - tan^4 = tan^4 sec^2 - tan^2 sec^2 + tan^2 = tan^4 sec^2 - tan^2 sec^2 + sec^2 - 1 now d(tanx) = sec^2x dx, so what you have is u^4 du - u^2 du + du - dx = 1/5 u^5 - 1/3 u^3 + u - x = 1/5 tan^5x - 1/3 tan^3x + ...
*Sunday, January 8, 2012 at 5:56pm by Steve*

**Trig - tan 15° using composite argument?**

tan 15° tan (45°-30°) (tan 45° - tan 30°)/1+ tan 45°tan30° (1-√3/3)/(1+1√3/3) then i donnt what to do/ chancel out. Can someone finish it if i didnt get it wrong. thanks in advance
*Wednesday, November 12, 2008 at 7:58pm by Anonymous*

**Solving Trig equation**

you should recognize that they are using the formula tan(A+B) = (tanA + tanB)/(1 - tanAtanB) so (tanx + tan π/3)/(1 - tanxtan(π/3) = √3 becomes tan(x+π/3) = √3 , I know tan 60° or tan 240° = √3 OR tan π/3 or tan 4π/3 = √3 so x...
*Friday, April 27, 2012 at 11:41pm by Reiny*

**physics**

Ignoring ball and hoop diameter, just solve y(x) = -gsec^2(θ)/2v^2 x^2 + tanθ x + h when y(4.57) = 3.05 -9.8 sec^2θ/(2*8^2) * 4.57^2 + tanθ (4.57) + 2.48 = 3.05 -1.6 (1+tan^2θ) + 4.57 tanθ - 0.57 = 0 1.6 tan^2θ - 4.57 tanθ + 2.17 = 0 tan...
*Monday, June 24, 2013 at 9:16pm by Steve*

**calculus**

additional note: Since tan(x)=tan(x±π), you can calculate tan(x-π) instead of tan(x). This way, the cosine series converges more rapidly.
*Saturday, February 6, 2010 at 8:01pm by MathMate*

**Pre calculus**

Let the building height be Y Let the horizontal distance to the building be h. h tan 42 = a h tan 8 = b Y = a + b = h (tan42,+ tan 8) For the value of h, use h tan 8 = 10 m
*Sunday, July 15, 2012 at 8:26pm by drwls*

**Maths**

Since 540 = 3*180 and the period of tan(x) is 180 tan(540-x) = tan(-x) = -tan(x) = -1/5 However, to answer your question, yo need to go clockwise, since you have tan(-x).
*Tuesday, May 22, 2012 at 5:28am by Steve*

**Math - help really needed**

I'm really sorry. I don't want to sound stupid, but when I add tan^2x/tan^2x to 1/tan^2x I get 1+tan^2x/tan^2x and then the tans cancel out and I'm left with just one. Obviously, that's not correct, but I can't find my mistake.
*Tuesday, December 2, 2008 at 9:33pm by Rani*

**trig**

tan x = 13 cot x tan x = 13/tan x tan^2 x = 13 tan x = ±√13 x = 74.5º,105.5º, 254.5º, 285.5º
*Thursday, May 1, 2008 at 12:12am by Reiny*

**more trig.... how fun!!!!**

if you can't help me with my first question hopw you can help me with this one. sec(-x)/csc(-x)=tan(x) thanx to anyone who can help From the definition of the sec and csc functions, and the tan function, sec(-x)/csc(-x) = sin(-x)/cos(-x) = tan(-x) However, tan (x) does not ...
*Monday, November 6, 2006 at 10:23pm by berhana*

**Maths Calculus Derivatives**

Find the first derivative for the following functions 1) f(x) = sin(cos^2x) cos(sin^3x) 2) f(x) = ( tan 2x - tan x ) / ( 1 + tan x tan 2x ) 3) f(x) = sin { tan ( sqrt x^3 + 6 ) } 4) f(x) = {sec^2(100x) - tan^2(100x)} / x
*Saturday, October 22, 2011 at 7:31am by Yousef*

**Maths**

cos^2(360-A) = cos^2(A) tan^2(180+A) = tan^2(A) sec^2(180-A) = sec^2(A) sin^2(540+A) = sin^2(A) csc^2(-A) = csc^2(A) cot^2(90+A) = tan^2(A) so, using those simplifications, we have cos^2 * tan^2 * sec^2 = tan^2 sin^2 * csc^2 * tan^2 = tan^2 and the fraction is just 1 How far ...
*Thursday, May 2, 2013 at 11:18pm by Steve*

**Trig**

tan(a-b) = [ tan a - tan b)/(1 + tan a tan b) You better learn those trig identities.
*Tuesday, December 27, 2011 at 6:42pm by Damon*

**Calc**

Make a table where the first column is x, the second column is tan(3x), the third is tan(5x), and the last is tan(3x)/tan(5x). Set your calculator to radians, and calculate the values of tan(3x) and tan(5x) for x=0.1, 0.01, 0.001... From the values suggested by the last column...
*Sunday, September 26, 2010 at 3:58pm by MathMate*

**Trigonometry**

drwls thats wrong :O tan 7pie/12 = tan (3pie/12 + 4pie/12) *break it down* = tan (pie/4 + pie/3) *find tan pie/4 and tan pie/3 on unit circle* = 1 + square root of 3 *answer* thats the exact value =] -Cyborg03
*Thursday, September 13, 2007 at 12:40am by William (Cyborg03)*

**calculus**

Let's define I_{n} = Integral of tan^n(x)dx You can express I_{n} in terms of I_{n-2} as follows. We write the integrand as: tan^n(x) = sin^n(x)/cos^n(x) Take out a factor sin^2(x): tan^n(x) = sin^2(x)sin^(n-2)/cos^n(x) And then substitute: sin^2(x) = 1 - cos^2(x) tan^n(x) = ...
*Thursday, September 25, 2008 at 6:50pm by Count Iblis*

**trig**

H = Hight of lighthouse L = Distance between a ship and a lighthouse tan( theta ) = H / L If 19^ mean 19° then: tan( theta ) = H / L tan( 19° ) = 120 / L L * tan( 19° ) = 120 Divide both sides with tan( 19° ) L = 120 / tan ( 19° ) L = 120 / 0.34433 L = 348.5 ft If 19^ mean 19...
*Saturday, October 1, 2011 at 11:19am by Anonymous*

**Calculus**

f'(x) = (1/2)(tan(1-x))^(-1/2)*sec^2 (1-x)(-1) = -(1/2) [tan(1-x)]^(-1/2) [sec(1-x)]^2 now we need the product rule, what a mess ... f''(x) = -(1/2) [tan(1-x)]^(-1/2) (2)(sec(1-x)) (sec(1-x))(tan(1-x))(-1) + (-1/2)[sec(1-x)]^2 (-1/2)tan(1-x))^(-3/2)(sec^2(1-x))(-1) check this ...
*Tuesday, April 13, 2010 at 4:01pm by Reiny*

**trigo math**

tan*sin+cos = sin^2/cos + cos = (sin^2+cos^2)/cos = 1/cos = sec (tan*cos^2 + sin^2)/sin = (sin*cos + sin^2)/sin = sin(cos+sin)/sin = cos+sin I think you mean (1+tan)/(1-tan) = (1+tan)^2/(1-tan^2) = (1+2tan+tan^2)/(1-tan^2) = (sec^2+2tan)/(1-tan^2) the last one needs some ...
*Thursday, December 13, 2012 at 10:58am by Steve*

**calcus**

integral (sec u) du This is pretty tricky, but what we'll do here is multiply both numerator and denominator by (sec u + tan u): = integral (sec u * (sec u + tan u)/(sec u + tan u)) du = integral (sec^2 (u) + sec(u)*tan(u))/(sec u + tan u)) du Then we use substitution. Let v...
*Tuesday, October 8, 2013 at 11:49am by Jai*

**Pre-Calc Help!!!**

LS = (tanx - tan(π/4)/(1 + tanx(tan(π/4)) ) = (tanx - 1)/(1 + tanx) = RS I used the identity tan(A-B)= (tanA - tanB)/(1 + tanAtanB) and the fact that tan π/4 = tan 45°= 1
*Friday, November 22, 2013 at 2:45pm by Reiny*

**Pre-Cal**

from you second line tan^2(x)/tan(x) - sec^2x/tanx = (tan^2 x - sec^2 x)/tanx = (tan^2 x - (1 + tan^2 x))/tanx = -1/tanx or -cotx or -sinx/cosx
*Friday, March 5, 2010 at 6:31pm by Reiny*

**calculus**

Yo can find a reduction formula for the integral of tan^n(x) as follows. We have that tan^n(x) = sin^n(x)/cos^n(x) = sin^(n-2)(x)/cos^n(x) sin^2(x) = sin^(n-2)(x)/cos^n(x) [1-cos^2(x)] = sin^(n-2)(x)/cos^n(x) - sin^(n-2)(x)/cos^(n-2)(x) = sin^(n-2)(x)/cos^(n-2)(x) 1/cos^2(x...
*Thursday, June 3, 2010 at 11:22pm by Count Iblis*

**trignonmetry**

PLEASE tell your classmates to use parentheses to clarify the problems! I think you mean (tan*cos^2 + sin^2)/sin = (sin*cos + sin^2)/sin = sin(cos+sin)/sin = cos+sin I think you mean (1+tan)/(1-tan) = (1+tan)^2/(1-tan^2) = (1+2tan+tan^2)/(1-tan^2) = (sec^2+2tan)/(1-tan^2) I ...
*Sunday, December 16, 2012 at 1:27pm by Steve*

**trig**

Draw a diagram. If the nearer point is a distance x from the base of the mountain, then the height h can be figured from h/x = tan 43° h/(x+235) = tan 30° x = h/tan 43° x = h/tan 30° - 235 h/tan 43° = h/tan 30° - 235 h = 235/(cot30° - cot43°) h = 356.2 m not much of a mountain.
*Wednesday, March 13, 2013 at 10:45pm by Steve*

**trig**

looks like you mean y = tan(2x + pi/4) = (tan 2x + tan pi/4)/(1 - (tan 2x)(tan pi/4) = (tan 2x + 1)/(1 - tan 2x), since tan pi/4 = 1
*Friday, March 5, 2010 at 4:11pm by Reiny*

**maths**

(1+tan x)/(1-tan x) + (1-tan x)/(1+tan x) = [(1+tan x)^2 + (1-tan x)^2]/(1-tan^2 x) = (2 + 2tan^2 x)/(1-tan^2 x) = 2(cos^2 x + sin^2 x)/(cos^2 x - sin^2 x) = 2/cos 2x = 2sec 2x
*Friday, August 10, 2012 at 5:09am by Steve*

**Trig**

s is in Q1 (sin>0, cos>0) so cos(s)=1/3, sin(s)=+√(1-(1/3)²) =√(8/9) tan(s)=√(8/9) / (1/3) =√8 t is in Q4 (sin<0, cos>0) sin(t)=-1/2 cos(t)=+√(1-(1/2)²) =√(3/4) tan(t)=(-1/2)/√(3/4) =-1/√3 [rationalize by ...
*Saturday, February 19, 2011 at 11:21pm by MathMate*

**Math - help really needed**

I'll give you a hint: Try replacing the 1 with tan^2x/tan^2x. After add that to 1/tan^2x, replace the denominator (the tan^2x) with sin^2/cos^2.
*Tuesday, December 2, 2008 at 9:33pm by Bob*

**Math **

Determine each tangent ratio to four decimal places using a calculator. A) tan 74 degrees B) tan 45 degrees C) tan 60 degrees D) tan 89 degrees How do I figure this out !?
*Thursday, January 19, 2012 at 10:19am by Courtney*

**Trigonometry**

Verify the identity algebraically. This problem is very intriguing and awesome at the same time. It's wonderfully amazing! 1.) TAN³α-1/TAN α-1= TAN²α + TAN α + 1
*Sunday, February 17, 2013 at 1:57am by AwesomeGuy*

**Maths- complex numbers**

Find tan(3 theta) in terms of tan theta Use the formula tan (a + b) = (tan a + tan b)/[1 - tan a tan b) in two steps. First, let a = b = theta and get a formula for tan (2 theta). tan (2 theta) = 2 tan theta/[(1 - tan theta)^2] Then write down the equation for tan (2 theta + ...
*Tuesday, February 13, 2007 at 2:45am by Jake*

**trig**

cos 2t = 2cos^2(t) - 1 = 2/sec^2(t) - 1 sec^2 = 1 + tan^2 2/(1 + tan^2(t)) - 1 = [2 - (1 + tan^2 t)]/(1 + tan^2 t) x = 4tan t tan t = x/4 [2 - ( 1 + x^2 / 16)]/(1 + x^2 / 16) (32 - 1 - x^2)/(16 + x^2) (31 - x^2)/(16 + x^2)
*Monday, October 17, 2011 at 10:21pm by Steve*

**Precalculus**

You have (tan+sec-1)/(tan-(sec-1)) multiply top and bottom by tan+sec-1 and you have (tan+sec-1)^2/(tan^2-(sec-1)^2) tan^2+sec^2+1+2tan*sec-2tan-2sec)/(tan^2-sec^2+2sec-1) (sec^2+tan*sec-tan-sec)/(sec-1) (tan+sec)(sec-1)/(sec-1) tan+sec
*Wednesday, April 16, 2014 at 2:46pm by Steve*

**Math - help really needed**

You actually get (1/tan^2x)+1, which is cot^2 + (csc^2x-cot^2x), which is csc^2x, which is 1/sin^2x. And that's your answer. (You could only cancel out the tan^2 if it were (Some #)(tan^2x)/tan^2x, rather than (1)+(tan^2x)/tan^2x. Dividing is the inverse of multiplying, so ...
*Tuesday, December 2, 2008 at 9:33pm by Bob*

**calculus **

So I am suppose to evaulate this problem y=tan^4(2x) and I am confused. my friend did this : 3 tan ^4 (2x) d sec^ 2x (2x)= 6 tan ^4 (2x) d sec^2 (2x) She says it's right but what confuses me is she deriving the 4 and made it a three? I did the problem like this: tan^4 (2x)= 4 ...
*Saturday, July 28, 2012 at 9:23pm by Anonymous*

**Trigonometry**

1.Solve tan^2x + tan x – 1 = 0 for the principal value(s) to two decimal places. 6.Prove that tan y cos^2 y + sin^2y/sin y = cos y + sin y 10.Prove that 1+tanθ/1-tanθ = sec^2θ+2tanθ/1-tan^2θ 17.Prove that sin^2w-cos^2w/tan w sin w + cos w tan w = cos ...
*Tuesday, January 1, 2013 at 12:51pm by Alonso*

**trig**

Given that tan 45=1, use tan(x+y) to show that tan 22.5= /2 - 1. (/is square root sign
*Tuesday, August 24, 2010 at 3:40am by Mimi*

**calculus**

Differentiate e^(2x^3+x) X tan(x^2) tan e^x / ln(x^6/2-3x+3e) cos(tan(e^4x^2))
*Sunday, March 30, 2008 at 1:43am by daria*

**math**

tan^-1a+tan^-1b+tan^-1c=pi,yhen prove that a+b+c=abc
*Wednesday, April 3, 2013 at 12:12pm by hasee*

**Math (trigonometry)**

*Sorry, should be: I'm just wondering...why did you write (1 + tan x) / x on the *right* side? Instead of (1 + tan x) / tan x?
*Sunday, April 13, 2008 at 9:53pm by Lucy*

**physics**

tan^-1(Vz/Vx) = tan^-1(az/ax) = tan^-1(5.9/3.2) = 61.5 degrees The angle will be independent of time.
*Thursday, June 23, 2011 at 12:10am by drwls*

**Amaths**

tan(A+B) = (tanA+tanB)/(1-tanAtanB) tan(A-B) = (tanA-tanB)/(1+tanAtanB) so, tanA+tanB = tan(A+B)(1-tanAtanB) tanA-tanB = tan(A-B)(1+tanAtanB) divide the two and you have tan(A+B)/tan(A-B) * (1-tanAtanB)/(1+tanAtanB) multiply top and bottom by cosAcosB nad you have tan(A+B)/tan...
*Saturday, January 11, 2014 at 3:07am by Steve*

**PreCalculus**

Tan(theta+n*PI)=tan(theta) tangent is periodic every pi radians. Tan(theta+n*180)=tan theta -tan380=-tan(380-180)=-tan200 Learn quickly those periodic formulas, or as I do, memorize the sine and cosine curves.
*Sunday, November 15, 2009 at 11:26pm by bobpursley*

**Pre-Calculus**

Ok i understand until you say that tan(theta) +tan(beta) needs to be divided by 1-tan(theta)tan(beta) where does that come from?
*Sunday, November 14, 2010 at 10:00pm by Riley*

**Math- Precalculus**

Write the expression as the sine, cosine, tangent of an angle (in radians.) (tan(pi/5)-tan(pi/3))/(1+tan(pi/5)tan(pi/3))
*Monday, May 20, 2013 at 2:19pm by Maddy*

**trig**

Use a sum or difference identity to find the exact value. tan 25deg + tan 5deg / 1-tan 25deg tan 5deg
*Monday, June 17, 2013 at 12:40am by tony*

**Trig**

Sum and diffence formula Finding Exact value of Tan 105-Tan 10)-15)/1+ tan(105)Tan(-15)
*Monday, December 9, 2013 at 8:59pm by Jacob*

**Calculus**

Wow. Looks good, but I'm stumped on the transformations to use. If you plot them at wolframalpha.com via plot log[sec arcsin {1/3*tan(x/2)}+ tan arcsin {1/3*tan(x/2)}] , arctanh {1/3*tan(x/2)} The two graphs overlap completely, so they must be equal.
*Wednesday, October 30, 2013 at 4:42am by Steve*

**trigonometry**

recalling that sec^2 = 1+tan^2, rearrange things to have 3tan^2 y + 5 tan y - 2 = 0 (3tan y - 1)(tan y + 2) = 0 tan y = 1/3 or -2 tan is positive in QI and QIII, so tany = 1/3 means that y = 18.43° or 198.43° tany = -2 means y = 116.57° or 296.57°
*Wednesday, May 2, 2012 at 4:15pm by Steve*

**Calc**

y= cube root (1+tan(t)) OR y= (1+tan(t))^(1/3) The answer I got was y'= (1/3)(1+tan(t))^(-2/3)*(sec^2(t)) Is this correct?
*Tuesday, October 19, 2010 at 3:06pm by Checking answer*

**Calculus : Derivative**

If tan^-1(x^2-y^2/x^2+y^2) = a, prove that dy/dx = x(1 - tan a)/y(1 + tan a) Please solve!!!
*Wednesday, June 15, 2011 at 3:06pm by Akash*

**Trigonometry Help**

divide by 4 and multiply by tan x tan^2 x - 3 = 0 tan x = ±√3 x = kpi ± pi/3 for integer k
*Sunday, November 18, 2012 at 8:06pm by Steve*

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