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प्रश्न
Prove that tan2Φ + cot2Φ + 2 = sec2Φ.cosec2Φ.
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उत्तर
L.H.S. = tan2Φ + cot2Φ + 2
= tan2Φ + 1 + cot2Φ + 1
= sec2Φ + cosec2Φ
= `1/cos^2 Φ + 1/sin^2Φ`
= `(sin^2 Φ + cos^2 Φ)/(sin^2 Φ.cos^2Φ )`
= `1/(sin^2 Φ. cos^2 Φ )`
= cosec2Φ. sec2Φ
= R.H.S.
Hence proved.
संबंधित प्रश्न
Prove the following identities:
`sqrt((1 - cosA)/(1 + cosA)) = cosec A - cot A`
Write the value of `4 tan^2 theta - 4/ cos^2 theta`
Find the value of ` ( sin 50°)/(cos 40°)+ (cosec 40°)/(sec 50°) - 4 cos 50° cosec 40 °`
Prove the following identity :
`(sec^2θ - sin^2θ)/tan^2θ = cosec^2θ - cos^2θ`
Find x , if `cos(2x - 6) = cos^2 30^circ - cos^2 60^circ`
Prove that `sin A/(sec A + tan A - 1) + cos A/(cosec A + cot A - 1) = 1`.
If sec θ + tan θ = `sqrt(3)`, complete the activity to find the value of sec θ – tan θ
Activity:
`square` = 1 + tan2θ ......[Fundamental trigonometric identity]
`square` – tan2θ = 1
(sec θ + tan θ) . (sec θ – tan θ) = `square`
`sqrt(3)*(sectheta - tan theta)` = 1
(sec θ – tan θ) = `square`
Prove that sec2θ + cosec2θ = sec2θ × cosec2θ
Prove the following:
`sintheta/(1 + cos theta) + (1 + cos theta)/sintheta` = 2cosecθ
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`
