For all 0 < θ < 9 0 ∘ ,
sec 2 θ + csc 2 θ = ?
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How 2 became 2tan cot
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If you convert tan^2 and cot^2 to sin^2/cos^2 and cos^2/sin^2 you can multiply (sin^2/cos^2)(sin^2/sin^2), (cos^2/sin^2)(cos^2/cos^2) and then 2(cos^2)(sin^2)/cos^2sin^2 and you get = (sin^4+2sin^2cos^2+cos^4)/sin^2cos^2. if you use algebra (x^4+2xy+y^4)= (x^2+y^2)(x^2+y^2) or sin^4...= (sin^2+cos^2)(sin^2+cos^2)/sin^2cos^2 and using trig identities we know that (sin^2+cos^2)=1, so your new answer is 1/sin^2cos^2. sqrt 1/sin^2cos^2= 1/sincos. then we turn the 1 back into sin^2+cos^2 and we get (sin^2/sincos)+(cos^2/sincos) or sin/cos+cos/sin which again converts to tan+cot.
Cot(x). Tan(x) = 1
cos 2 x 1 + sin 2 x 1
sin 2 x × cos 2 x cos 2 x + sin 2 x
sin x cos x 1
sin x cos x cos 2 x + sin 2 x
= tan x + cot x
Okay, thanks!
you confuse me
nevermind i understood
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sec 2 θ + csc 2 θ
= ( 1 + tan 2 θ ) + ( 1 + cot 2 θ )
= tan 2 θ + 2 + cot 2 θ
= tan 2 θ + 2 tan θ cot θ + cot 2 θ
= ( tan θ + cot θ ) 2
tan θ + cot θ