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प्रश्न
Prove that : `tan"A"/(1 - cot"A") + cot"A"/(1 - tan"A") = sec"A".cosec"A" + 1`.
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उत्तर
LHS = `(sin"A"/cos"A")/(1 - cos"A"/sin"A") + (cos"A"/sin"A")/(1 - sin"A"/cos"A")`
= `(sin"A" sin"A")/(cos"A"(sin"A" - cos"A")) + (cos"A" cos"A")/((cos"A" - sin"A") sin"A"`
= `1/((sin"A" - cos"A")) [(sin^2"A")/cos"A" + (cos^2"A")/(-sin"A")]`
= `(sin^3"A" - cos^3"A")/(sin"A".cos"A"(sin"A" - cos"A"))`
= `((sin"A" - cos"A")(sin^2"A" + cos^2"A" + sin"A". cos"A"))/(sin"A". cos"A"(sin"A" - cos"A")`
= `(1 + sin"A". cos"A")/(sin"A".cos"A")`
= `1/(sin"A".cos"A") + (sin"A".cos"A")/(sin"A".cos"A")`
= `1/sin"A" . 1/cos"A"+ 1`
= sec A.cosec A + 1
= RHS
Hence proved.
संबंधित प्रश्न
Prove that
`sqrt((1 + sin θ)/(1 - sin θ)) + sqrt((1 - sin θ)/(1 + sin θ)) = 2 sec θ`
Prove the following identities:
`(1 + sin A)/(1 - sin A) = (cosec A + 1)/(cosec A - 1)`
Prove the following identities:
(1 – tan A)2 + (1 + tan A)2 = 2 sec2A
Prove the following identities:
`(sinAtanA)/(1 - cosA) = 1 + secA`
If 2 sin A – 1 = 0, show that: sin 3A = 3 sin A – 4 sin3 A
Prove the following identity :
`tanA - cotA = (1 - 2cos^2A)/(sinAcosA)`
If (sin α + cosec α)2 + (cos α + sec α)2 = k + tan2α + cot2α, then the value of k is equal to
Prove the following identity:
(sin2θ – 1)(tan2θ + 1) + 1 = 0
`1/sin^2θ - 1/cos^2θ - 1/tan^2θ - 1/cot^2θ - 1/sec^2θ - 1/("cosec"^2θ) = -3`, then find the value of θ.
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`
