Question

ABC is a right-angled triangle, right-angled at B. Given that ∠ACB = θ, side AB = 2 units and side BC = 1 unit, find the value of sin²θ + tan²θ.

Answer 1

Given: ABC is right-angled at B, ∠ACB = θ, AB = 2, BC = 1.

Step-wise calculation:

1. Hypotenuse `AC = sqrt(AB^2 + BC^2)` 

= `sqrt(2^2 + 1^2)`

= `sqrt(5)`

2. `sin θ = "Opposite"/"Hypotenuse"`

= `"AB"/"AC"`

= `2/sqrt(5) . cos θ`

= `"Adjacent"/"Hypotenuse"`

= `"BC"/"AC"`

= `1/sqrt(5) . tan θ`

= `"Opposite"/"Adjacent"`

= `"AB"/"BC"`

= 2

3. If the intended expression is sin²θ + tan²θ:

`sin^2θ = (2/sqrt(5))^2`

= `4/5`

tan²θ = 2² 

= 4

= `20/5`

Sum = `4/5 + 20/5`

= `24/5`

4. If instead the intended expression was sin (2θ) + tan (2θ)):

sin (2θ) = 2 sin θ cos θ

= `2 xx (2/sqrt(5)) xx (1/sqrt(5))`

= `4/5`

tan (2θ) = `(2 tan θ)/(1 - tan^2θ)`

= `2 xx 2/(1 - 4)`

= `4/(-3)`

= `-4/3`

Sum = `4/5 - 4/3`

= `(12 - 20)/15`

= `-8/15`

Interpreting sin2θ + tan2θ as sin²θ + tan²θ: value = `24/5`.

Interpreting sin2θ + tan2θ as sin(2θ) + tan(2θ): value = `-8/15`.