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# Prove tanx + cotx = secxcscx?

could someone explain how you would answer this question.

### 8 Answers

- 1 decade agoFavorite Answer
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

=]

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- workoutgirl1235Lv 61 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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- bskelkarLv 71 decade ago
LHS = [sinx/cosx] + [cosx/sinx]

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

=1/(cosxsinx)

= (1/sinx)(1/cosx)

= (cossecx)(secx)

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- HARMEET SINGHLv 51 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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- ledfordLv 43 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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