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प्रश्न
`(1 - tan^2 45^circ)/(1 + tan^2 45^circ)` = ?
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उत्तर
`(1 - tan^2 45^circ)/(1 + tan^2 45^circ)` = `(1 - (1)^2)/(1 + (1)^2)` ......[∵ tan 45° = 1]
= `(1 - 1)/(1 + 1)`
= `0/2`
= 0
संबंधित प्रश्न
Prove the following identities, where the angles involved are acute angles for which the expressions are defined:
`sqrt((1+sinA)/(1-sinA)) = secA + tanA`
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`cos^2 A + 1/(1 + cot^2 A) = 1`
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sin2 A cot2 A + cos2 A tan2 A = 1
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`1 - sin^2A/(1 + cosA) = cosA`
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Write the value of `(sin^2 theta 1/(1+tan^2 theta))`.
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tanA+cotA=secAcosecA
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If tan θ = `9/40`, complete the activity to find the value of sec θ.
Activity:
sec2θ = 1 + `square` ......[Fundamental trigonometric identity]
sec2θ = 1 + `square^2`
sec2θ = 1 + `square`
sec θ = `square`
Prove that sec2θ – cos2θ = tan2θ + sin2θ
Proved that `(1 + secA)/secA = (sin^2A)/(1 - cos A)`.
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`
