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Question
Prove that `sqrt(2 + tan^2 θ + cot^2 θ) = tan θ + cot θ`.
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Solution
LHS = `sqrt(2 + tan^2 θ + cot^2 θ)`
= `sqrt( tan^2 θ + cot^2θ + 2tan θ.cot θ)` ...[ ∵ tan θ.cot θ = 1 ]
= `sqrt( tan^2 θ + cot^2θ)`
= tan θ + cot θ
= RHS
Hence proved.
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Find the value of sin2θ + cos2θ
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`
