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Question
Prove the following identities:
cosec A(1 + cos A) (cosec A – cot A) = 1
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Solution
L.H.S. = cosec A(1 + cos A) (cosecA – cot A)
= `1/(sin A)(1 + cos A)(1/(sin A) - (cos A)/(sin A))`
`((1-cos A)/sin A)`
`1/sin A(1+cos A)((1-cos A)/sin A)`
`= ((1+ cos A)(1-cos A))/sin^2 A`
Apply the identity (1 + cosA) (1 − cosA) = 1 − cos2A
`= (1-cos^2A)/sin^2A`
`= sin^2A/sin^2A = 1`
cscA(1 + cosA) (cscA − cotA) = 1 proved
RELATED QUESTIONS
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`(cot A + tan B)/(cot B + tan A) = cot A tan B`
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`(secA - tanA)/(secA + tanA) = 1 - 2secAtanA + 2tan^2A`
If tan A = n tan B and sin A = m sin B, prove that `cos^2A = (m^2 - 1)/(n^2 - 1)`
`(1+ tan theta + cot theta )(sintheta - cos theta) = ((sec theta)/ (cosec^2 theta)-( cosec theta)/(sec^2 theta))`
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Eliminate θ if x = r cosθ and y = r sinθ.
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`
