Question

If sin θ = `12/13`, Find `(sin^2 θ - cos^2 θ)/(2sin θ cos θ) × 1/(tan^2 θ)`.

Answer 1

Given: Sin θ = `12/13 = "AB"/"AC"`

Let, AB = 12k and AC = 13k

In ΔABC, ∠B = 90°
By pythagoras theorem,
AB² + BC² = AC²
(12k)² + BC² = (13k)²
144k² + BC² = 169k²
BC² = 169k² - 144k²
BC² = 25k²
Taking square root,
BC = 5k
∴ Cos θ = `"BC"/"AC" = "5k"/"13k" = 5/13`

∴  tan θ = `"AB"/"BC" = "12k"/"5k" = 12/5`

Now, 

`(sin^2 θ - cos^2 θ)/(2sin θ cos θ) × 1/(tan^2 θ)`.

⇒ `[(12/13)^2 - (5/13)^2]/[2 × 12/13 × 5/13] × 1/(12/5)^2`

⇒ `[(144/169) - (25/169)]/[120/169] × 1/(144/25)`

⇒ `[(144/169) - (25/169)]/[120/169] × 25/144`

⇒ `((144 - 25)/cancel169)/[120/cancel169] × 25/144`

⇒ `119/120 × 25/144`

⇒ `595/3456`