Question

If tan α = 2, evaluate sin α sec α + tan²α – cosec α.

Answer 1

Given: tan α = 2.

Step-wise calculation:

1. `sin α xx sec α = sin α xx (1/cos α)`

= tan α

= 2

2. tan²α = 2²

= 4

3. `"cosec"  α = 1/(sin α)`.

From `tan α = 2 = (sin α)/(cos α)` set sin α = 2k, cos α = k. 

Then (2k)² + k² = 1

⇒ 5k² = 1

⇒ `k = ±1/sqrt(5)`

If α is acute (sin α > 0), sin α = `2/sqrt(5)` so cosec α = `sqrt(5)/2`.

Combine: sin α sec α + tan²α − cosec α

= `2 + 4 - (sqrt(5)/2)` 

= `6 - sqrt(5)/2` 

= `(12 - sqrt(5))/2`

Assuming α is acute, the value is `(12 - sqrt(5))/2`. If α were in a quadrant where sin α < 0, the result would be `(12 + sqrt(5))/2`.