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Question
If sec θ + tan θ = x, write the value of sec θ − tan θ in terms of x.
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Solution
Given:` secθ+tanθ=x`
We know that,
`Sec^2θ-tan^2θ=1`
Therefore,
`sec^2 θ-tan^2θ=1`
⇒` (Secθ+tan θ) (Secθ-tan θ)=1`
⇒` x (secθ-tan θ )=1`
⇒ `(sec θ-tan θ)=1/x`
Hence, `sec θ-tan θ=1/4`
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Solution:
In Δ ABC, ∠ABC = 90°, ∠C = θ°
AB2 + BC2 = `square` .....(Pythagoras theorem)
Divide both sides by AC2
`"AB"^2/"AC"^2 + "BC"^2/"AC"^2 = "AC"^2/"AC"^2`
∴ `("AB"^2/"AC"^2) + ("BC"^2/"AC"^2) = 1`
But `"AB"/"AC" = square and "BC"/"AC" = square`
∴ `sin^2 theta + cos^2 theta = square`
Statement 1: sin2θ + cos2θ = 1
Statement 2: cosec2θ + cot2θ = 1
Which of the following is valid?
