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Question
Prove the following identity :
`1/(cosA + sinA - 1) + 2/(cosA + sinA + 1) = cosecA + secA`
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Solution
LHS = `1/((cosA + sinA) - 1) + 1/((cosA + sinA) + 1)`
= `(cosA + sinA + 1 + cosA + sinA - 1)/((cosA + sinA)^2 -1)`
= `(2(cosA + sinA))/(cos^2A + sin^2A + 2cosAsinA - 1)`
= `(2(cosA + sinA))/(1 + 2cosAsinA - 1) = (cosA + sinA)/(cosAsinA)`
= `cosA/(cosAsinA) + sinA/(cosAsinA)`
= `1/sinA + 1/cosA`
= cosecA + secA
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RELATED QUESTIONS
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`(cot^2 theta ( sec theta - 1))/((1+ sin theta))+ (sec^2 theta(sin theta-1))/((1+ sec theta))=0`
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`(1 + cosA)/(1 - cosA) = (cosecA + cotA)^2`
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If `sec θ + tan θ = sqrt(3)`, complete the activity to find the value of sec θ – tan θ.
Activity:
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Prove the following:
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(1 – cos2 A) is equal to ______.
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`
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But `"AB"/"AC" = square and "BC"/"AC" = square`
∴ `sin^2 theta + cos^2 theta = square`
