Prove tanx + cotx = secxcscx?

could someone explain how you would answer this question.

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  • 1 decade ago
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    tan x = (sin x)/(cos x)

    cot x = (cos x)/(sin x)

    So you have:

    (sin x)/(cos x) + (cos x)/(sin x)

    Find a common denominator between the two to add them. The common denominator is (cos x)(sin x). Multiply the top and bottom of the first term by (sin x). Multiply the top and bottom of the second term by (cos x).

    You get:

    [(sin x)^2 + (cos x)^2] / [(cos x)(sin x)]

    and since (sin x)^2 + (cos x)^2 = 1,

    So you get 1/[(cos x)(sin x)]

    Which is (sec x)(csc x)

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  • 1 decade ago

    change tanx to sinx/cosx then change cotx to cosx/sinx

    change secx to 1/cosx and cscx to 1/sinx

    sinx/cosx + cosx/sinx=1/cosx(1/sinx)

    when u add u add or subtract u need to get common denominators for that side of the equation so times sinx/cosx by sinx and times cosx/sinx by cosx

    sin^2x +cos^2x/cosxsinx=1/cosx(1/sinx)

    sin^2x + cos^2x = 1 so substitute it for 1

    1/cosxsinx=1/cosx(1/sinx)

    so now for the other side of the equation multiple across with 1/cosx(1/sinx) and u get.....

    1/cosxsinx=1/cosxsinx

    =]

    Source(s): me =]
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  • 1 decade ago

    tanx + cotx = secxcscx

    sinx/cosx + cosx/sinx =

    sin²x/(sinxcosx) + cos²x/(sinxcosx) =

    (sin²x + cos²x)/(sinxcosx) =

    1/(sinxcosx) = secxcscx

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  • 1 decade ago

    ok so the equation is the same as sinx/cosx + cosx/sinx = (1/cosx)(1/sinx)

    multiply each side by sinx

    sin^2x/cosx + cosx = (1/cosx)

    then multiply each side by cosx

    sin^2x + cos^2x = 1

    then use one of the pathagorean identities cos^2x=1-sin^2x

    sin^2x + 1 - sin^2x = 1

    simplify the sin^2x and you get

    1 = 1

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  • 1 decade ago

    LHS = [sinx/cosx] + [cosx/sinx]

    = [(sinx)^2+(cosx)^2]/(cosxsinx)

    =1/(cosxsinx)

    = (1/sinx)(1/cosx)

    = (cossecx)(secx)

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  • 1 decade ago

    taking left hand side we have

    tanx + cotx=sinx/cosx+cosx/sinx

    = (sinx)^2+(cosx)^2/sinxcosx...........................(sinx)^2+(cosx)^2=1

    =1/sinxcosx=secxcscx=right hand side

    i hope this helps thanks and may god help u dear!!!

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  • 3 years ago

    tan x = sin x / cos x cot x = one million/tan x = cos x /sin x So, tan x + cot x = (sin x/cos x) + (cos x/sin x) = (sin^2 x + cos^2 x)/(sin x cos x) sin^2 x + cos ^ x = one million So, tan x + cot x = one million/(sin x cos x) sec x = one million/cos x csc x = one million/sin x So, sec x csc x = (one million/cos x)(one million/sin x) = one million/(cos x sin x)

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  • 1 decade ago

    sinx/cosx+cosx/sinx

    (sin^2x+cos^2x)/(cosxsinx)

    1/(cosxsinx)

    secxcscx

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