# How do you prove the identity: `(tanx+cotx)^2=sec^2x csc^2x ?` ``

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### 3 Answers

`(tanx + cotx)^2 = sec^2 x csc^2 x`

`tan^2 x + 2tanxcotx + cot^2 x = sec^2 x csc^2 x`

`(sinx/cosx)^2 + 2(sinx/cosx)(cosx/tanx) + (cosx/sinx)^2 = sec^2 x csc^2 x`

`==> (sin^2 x)/(cos^2 x) + 2(1) + (cos^2 x)/(sin^2 x) =sec^2 x csc^2 x`

`==> 2 + (sin^4 x + cos^4 x)/(sin^2x cos^2 x)= sec^2 x csc^2 x`

`==> (2sin^2 x cos^2 x + sin^4 x + cos^4 x)/(sin^2 x cos^2 x)= sec^2 x csc^2 x`

`==> (sin^2 x + cos^2 x)^2 /(sin^2 x cos^2 x)= sec^2 x csc^2 `

`==> 1/(sin^2 x cos^2 x)= sec^2 x csc^2 x`

`==> 1/(sin^2 x) X 1/(cos^2 x) = sec^2 x csc^2 x`

`==> sec^2 x cos^2 x = sec^2 x csc^2 x`

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We have to prove: `(tan x + cot x)^2 = sec^2 x + cosec^2x`

`sec^2x + cosec^2x`

=> `1/(cos^2 x) + 1/(sin^2 x)`

=> `(sin^2x + cos^2x)/(sin^2x*cos^2x)`

=> `1/(sin^2x*cos^2x)` ...(1)

`(tan x + cot x)^2`

=> `tan^2 x + cot^2x + 2*tan x*cot x`

=> `(sin^2x)/(cos^2x) + (cos^2x)/(sin^2x) + 2`

=> `(sin^4x + cos^4x)/(cos^2x*sin^2x) + 2`

=> `((sin^2x + cos^2x)^2 - 2*sin^2x*cos^2x + 2*sin^2x*cos^2x)/(cos^2x*sin^2x)`

=> `1/(cos^2x*sin^2x)` ...(2)

**As (1) = (2), the identity is proved.**

(tan x+cotx)^2 = (sinx/cosx + cosx/sinx)^2

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

=(1/cosx. sinx)^2 [since sin^2x + cos^x= 1]

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

=(cosec x sec x)^2

= cosex ^2 x . sec ^2x

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