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प्रश्न
If cotθ = `sqrt(7)`, show that `("cosec"^2θ -sec^2θ)/("cosec"^2θ + sec^2θ) = (3)/(4)`
उत्तर
cotθ = `sqrt(7)`
⇒ `cosθ/sinθ = sqrt(7)`
⇒ `"Base"/"Hypotenuse" xx "Hypotenuse"/"Perpendicular" = sqrt(7)/(1)`
⇒ `"Base"/"Perpendicular" = sqrt(7)/(1)`
Hypotenuse
= `sqrt(("Perpendicular")^2 + ("Base")^2`
= `sqrt(1 + 7)`
= `2sqrt(2)`
To show: `("cosec"^2θ -sec^2θ)/("cosec"^2θ + sec^2θ) = (3)/(4)`
`("cosec"^2θ -sec^2θ)/("cosec"^2θ + sec^2θ)`
= `(("Hypotenuse"/"Perpendicular")^2 - ("Hypotenuse"/"Base")^2)/(("Hypotenuse"/"Perpendicular")^2 + ("Hypotenuse"/"Base")^2)`
= `((2sqrt(2)/1)^2 - (2sqrt(2)/sqrt(7))^2)/(((2sqrt(2))/1)^2 + (2sqrt(2)/sqrt(7))^2`
= `(8/1 - 8/7)/(8/1 + 8/7)`
= `((56 - 8)/7)/((56 + 8)/7)`
= `(48)/(64)`
= `(3)/(4)`.
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