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