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Question
If cosec θ = 2x and \[5\left( x^2 - \frac{1}{x^2} \right)\] \[2\left( x^2 - \frac{1}{x^2} \right)\]
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Solution
Given:
`cosecθ=2x, cot θ2/x`
We know that,
`cosec^2 θ-cot^2 θ=1`
⇒` (2x)^2-(2/x)^2=1`
⇒` 4x^2-4/x^2=1`
⇒ `4(x^2-1/x^2)=1`
⇒`2xx2xx(x^2-1/x^2)=1`
⇒ `2(x^2-1/x^2)=1/2`
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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`
