$$2\cot4x = \cot2x \tan2x$$ 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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Int Cos8x 1 Tan2x Cot2x Dx Youtube
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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This is what I've got so far sin = 1/csc tan = 1/cot 1cot2(x) = csc2(x) 1/1cot2(x) = 1/csc2(x) 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^2x1 B 1/sec^2x 3 sec x/cscx C sin(x) 4 1sin^2x Dcsc^2xcot^2xsin^2x 5 cos^2 x E Prove the following identities $$(\sec^2 x \tan^2x)(\csc^2 x \cot^2x) = 1 2 \sec^2x \csc^2 x \tag i$$ $$\frac{\cos x}{1\tan x} \frac{\sin x}{1\cot x} = \sinMath please help quick Which of the following are identities?
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This question was just answered and I'm not sure if this is the same person or not If so, please let me know where you're stuck andSimplify (tan (2x))/ (1cot (2x)) tan (2x) 1 − cot(2x) tan ( 2 x) 1 cot ( 2 x) Nothing further can be done with this topic Please check the expression entered or try another topicGet stepbystep solutions from expert tutors as fast as 1530 minutes Your first 5 questions are on us!
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