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Questions
Prove the following trigonometric identities.
(sec2 θ − 1) (cosec2 θ − 1) = 1
Prove the following:
(sec2 θ − 1) (cosec2 θ − 1) = 1
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Solution
We know that
sec2 θ − tan2 θ = 1
cosec2 θ − cot2 θ = 1
So,
(sec2 θ − 1)(cosec2 θ − 1) = tan2 θ × cot2 θ
= (tan θ × cot θ)
= `(tan θ xx 1/tan θ)^2`
= (1)2
= 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`
