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Question
यदि `(sec α)/("cosec" β) = p` और `(tan α)/("cosec" β) = q`, है, तो सिद्ध कीजिए कि (p2 – q2) sec2α = p2
Theorem
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Solution
p = `(sec α)/("cosec" α)`
= `(1/(cos α))/(1/(sinα))`
= `(sin β)/(cos α)`
q = `(tan α)/("cosec" β)`
= `(sin α)/(cos α) 1/(sin β)`
= `(sin α)/(cos α) xx (sin β)/1`
= `(sin α . sin β)/(cos α)`
अब, (p2 – q2) sec2 α
= `((sin^2 β)/(cos^2 α) - (sin^2 α sin^2 β)/(cos^2 α)) sec^2 α`
अंश में sin2 β को सामान्य लें।
= `sin^2 β (1/(cos^2 α) - (sin^2 α)/(cos^2 α)) sec^2 α`
= sin2 β (sec2 α – tan2 α) . sec2 α
= `sin^2 β . 1/(cos^2 α)` (1) [∵ sec2 α – tan2 α = 1]
= `(sin^2 β)/(cos^2 α)`
= p2
∴ (p2 – q2) sec2 α = p2
अतः, सिद्ध हुआ।
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