Question

Prove that `(sin (90^circ - θ))/("cosec" (90^circ - θ)) + (cos (90^circ - θ))/(sec (90^circ - θ)) = 1`.

Answer 1

Given: Expression: `(sin (90^circ - θ))/("cosec" (90^circ - θ)) + (cos (90^circ - θ))/(sec (90^circ - θ))`

To Prove: The given expression equals 1.

Proof [Step-wise]: 

Step 1: Use complementary-angle identities:

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

Step 2: Use reciprocal identities for the co- and secants:

cosec (90° – θ) = sec θ and sec (90° – θ) = cosec θ

Step 3: Substitute these into the expression:

`sin(90^circ - θ)/("cosec" (90^circ - θ)) + cos(90^circ - θ)/(sec (90^circ - θ))` 

= `(cos θ)/(sec θ) + (sin θ)/("cosec"  θ)`

Step 4: Replace sec θ and cosec θ by their reciprocals `(sec θ = 1/(cos θ), "cosec"  θ = 1/(sin θ))`: 

`(cos θ)/(sec θ) = cos θ · cos θ` 

= cos²θ

`(sin θ)/("cosec"  θ) = sin θ · sin θ`

= sin²θ

Step 5: Add the results:

cos²θ + sin²θ = 1   ...(Pythagorean identity)

Therefore, the original expression equals 1.