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प्रश्न
sec2θ – tan2θ = ?
पर्याय
0
1
2
`sqrt(2)`
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उत्तर
1
Explanation:
1 + tan2θ = sec2θ
∵ sec2θ – tan2θ = 1
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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`
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