Question

Prove that `(tan θ)/(tan(90^circ - θ)) + (sin(90^circ - θ))/(cos θ) = sec^2θ`.

Answer 1

Given: Let θ be an angle.

To Prove: `(tan θ)/(tan(90^circ - θ)) + (sin(90^circ − θ))/(cos θ) = sec^2θ`.

Proof [Step-wise]:

1. Use cofunction identities:

tan (90° – θ) = cot θ and sin (90° – θ) = cos θ

2. Replace these into the left-hand side (LHS): 

LHS = `(tan θ)/(tan (90^circ - θ)) + (sin (90^circ - θ))/(cos θ)`

= `(tan θ)/(cot θ) + (cos θ)/(cos θ)`

3. Simplify each term:

`(tan θ)/(cot θ) = tan θ · tan θ`

= tan²θ

`(cos θ)/(cos θ) = 1` 

So, LHS = tan²θ + 1.

4. Use the Pythagorean identity 1 + tan²θ = sec²θ to get LHS = sec²θ.

Hence, `(tan θ)/(tan(90^circ - θ)) + (sin(90^circ − θ))/(cos θ) = sec^2θ`, as required.