# Verify the following identity: `tanx(cot x + tan x) = sec^2x`

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### 3 Answers

We have to verify that `tanx(cot x + tan x) = sec^2x`

Start with `tanx(cot x + tanx)`

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

=> `1 + tan^2x`

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

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

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

=> `sec^2x`

**This proves that `tan x(cot x + tan x) = sec^2 x` **

An alternative method is to take `sec^2 x ` and replace it by `1/(cos^2 x).`

Replace `1/(cos^2 x)` by `1+ tan^2 x` (the basic formula of trigonometry `1+tan^2 x = 1/(cos^2 x)` ).

Use this substitution you've made in the identity to be proven.

`tan x(cot x + tan x) = 1+tan^2 x`

Opening the brackets, you'll have`tanx*cotx + tan^2x = 1+tan^ 2 x`

The cotangent function is the inverse of tangent, therefore tanx*cot x = 1.

`1+ tan^2x = 1+tan^ 2 x`

**ANSWER: The last line proves the identity`tan x(cot x + tan x) = sec^2 x` **

L:H:S ≡ tanx(cotx + tanx)

= tanx.cotx + tan²x

**⇒ tanθ.cotθ = 1**

= 1 + tan²x

**⇒ 1 + tan²θ = sec²θ**

= sec²x

= R:H:S