70以上 1 tan^2x/1 cot^2x=tan^2x 324405-1-tan^2x/1-cot^2x=1-sec^2x

 $$2\cot⁡4x = \cot⁡2x \tan⁡2x$$ Thank you in advance Thank you for the comments and hints I got an answer after many tries ;) Below is my answer Thank you $2cot4x = cot2x tan2x$ $2\frac{1}{tan 4x} = cot2x tan2x$ $2\frac{1}{\frac{2 tan 2x}{1 tan^2 2x}} = cot2x tan2x$ $2\frac{1 tan^2 2x}{2 tan 2x} = cot2x tan2x$So sec^2 (x)=1tan^2 (x) This is one of the three Pythagorean identities in trigonometry, but if you don't recognize it, try converting to sines and cosines 1/cos^2 (x)=1sin^2 (x)/cos^2 (x) Now, multiply each term by cos^2 (x) to get 1=cos^2 (x) sin^2Found 2 solutions by ewatrrr, MathLover1 Answer by ewatrrr () ( Show Source ) You can put this solution on YOUR website!

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1-tan^2x/1-cot^2x=1-sec^2x

1-tan^2x/1-cot^2x=1-sec^2x-About Press Copyright Contact us Creators Advertise Developers Terms Privacy Policy & Safety How works Test new features Press Copyright Contact us Creators Calculus 2, integral of (1 tan^2x) sec^2x, integral of cos(2x) Hi simplifying the following (sec^2x csc^2x) (tan^2x cot^2x) tan^2x = sec^2x 1 cot^2x = csc^2x 1 (sec^2x csc^2x) (sec^2x 1 csc^2x 1)= 2click here👆to get an answer to your question ️ if sec x sec^ 2x = 1 then the value of tan^ 8 tan^ 4 2tan^ 2x 1 will be equal tox = 1287 2 = the period of the function is the

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$\begingroup$ Well, if your instructor insisted that you do this by calling in the doubleangle formula, then I would replace my second paragraph with a criticism of the instructor for making things unnecessarily difficult Indeed, if the question had been to solve $\tan(9x/2)=1$, it would have been frustratingly difficult to use the ninefold angle formula and the halfangle formulas,Trigo Identities We should know that 1 (tan^2x)=sec^2x also 1 (cot^2x)=csc^2x Substituting these identities to the eqn (tan^2x) (cot^2x)= (sec 1 tan2x= 1sin2x/cos2x Find common denominator which is cos2x cos2x/cos2x sin2x/cos2x (cos2x sin2x)/cos2x 1cot2x = 1 cos2x/sin2x Common denominator is sin2x sin2x/sin2x cos2x/sin2x ( sin2x cos2x)/sin2x 1/(cosx2 sin2x)/cos2x 1/( sin2x cos2x)/sin2x

See the answer See the answer See the answer done loading Show transcribed image text Expert Answer Who are the experts?Giải phương trình 1, \(2\tan^2x3\tan x 2\cot^2x3\cot x 2=0\) 2, \(\cos^23x\cos2x\cos^2x=0\) 3, \(\cos^22x2\left(\cos x \sin x\right)^23\sin2x 1=0\) 4, \(1Sin(a/2)= square root of (1cos(a))/2 Since the radical is on the right side of the equation , switch the sides so it is on the left side of the equation To remove the radical on the left side of the equation , square both sides of the equation

Mylove mylove Mathematics High School answered Cot^2xtan^2x=1 for all values of x true or false?Hi Simplifying the following (sec^2x csc^2x) (tan^2x cot^2x) tan^2x = sec^2x 1 cot^2x = csc^2x 1 (sec^2x csc^2x) (sec^2x 1 csc^2x 1)= 2Chứng minh đẳng thức (tan^3x/sin^2x)(1/sinxcosx) (cot^3x/cos^2x)=tan^3x cot^3x

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A=√sin2x(1 cotx) cos2x(1 tanx) B=sin^2xtan^2x/cos^2xcot^2x CẦN GẤP ẠAnswers Click here to see ALL problems on Trigonometrybasics Question tan^2xcot^2x=sec^2xcsc^2x Answer by jojo (1513) ( Show Source ) You can put this solution on YOUR website!Answers pineapplepizaaaaa Here for a good time of free question carrieaj08 no association stepbystep explanation

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Here, the problem is (cot^2 x 1)/ (tan^2 x 1) so cotx = 1/tanx by putting the value of cotx in above equation the simplified equation is (1 tan^2 x)/ (tan^2 x 1) (tan^2 x) (tan^2 x 1)/ (tan^2 x 1) (tan^2 x) 1/tan^2 x cot^2 x that should be the(tan x cot x)^2 = tan^2 x 2 tan x cot x = tan^2x cot^2x = (tan^2x 1) (cot^2 ) = This problem has been solved!Question prove the following identity cos2A = 2cos^2A 1 Answer by sarah_adam (1) ( Show Source ) You can put this solution on YOUR website!

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Cos2A = Cos (AA) we know the formula for Cos (AB)=CosACosBSinASinB therefore Cos (AA)= CosACosA SinASinA = Cos^2A Sin^2A WE also know that Cos^2A Sin^2A = 1Prove each identity a) 1cos^2x=tan^2xcos^2x b) cos^2x 2sin^2x1 = sin^2x I also tried a question on my own tan^2x = (1 – cos^2x)/cos^2x RS= sin^2x/cos^2x I know that the Pythagorean for that is sin^2x cos^2x That's all I could do Trigonometry Express sec2x in terms of tanx and secx I know you have to sec(2x) = 1/cos(2x) = 1/(cos²x sin²x) But how do you split that Like how to simplify that?

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Find The Derivative Of The Given Function Y Tan 2x 1 Cot 2x I Tried Converting The Original Function In Terms Of Sin And Cos But It Was Still Too Complicated To Be Called Simplified

 Find an answer to your question Cot^2xtan^2x=1 for all values of x true or false?Cot^2xcsc^2x=1 for all values of x true or falsse Question Cot^2xcsc^2x=1 for all values of x true or falsseExperts are tested by Chegg as specialists in their subject area We review their content and use

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Incoming Term: 1+tan^2x/1+cot^2x=tan^2x, 1-tan^2x/1-cot^2x=1-sec^2x, 1+tan^2x/1+cot^2 x=(1-tanx/1-cotx)^2,