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Question
Prove the following identities:
(sec A + cos A)(sec A − cos A) = tan2A + sin2A
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Solution
LHS = (sec A + cos A)(sec A − cos A)
LHS = sec2A − cos2A ...[(a + b)(a − b) = a2 − b2]
LHS = `1/cos^2"A" - cos^2"A"`
LHS = `(1 - cos^4"A")/cos^2"A"`
LHS = `(1^2-(cos^2"A")^2)/(cos^2"A")`
LHS = `((1 - cos^2"A")(1 + cos^2"A"))/cos^2"A" ...[a^2 − b^2 = (a + b)(a − b)]`
LHS = `(sin^2"A"(1 + cos^2"A"))/cos^2"A"`
LHS = `(sin^2"A" + sin^2"A"cos^2"A")/cos^2"A"`
LHS = `(sin^2"A")/(cos^2"A") + (sin^2"A"cos^2"A")/(cos^2"A")`
LHS = `sin^2"A"/cos^2"A" + sin^2"A"`
LHS = tan2A + sin2A
RHS = tan2A + sin2A
∴ LHS = RHS
∴ (sec A + cos A)(sec A − cos A) = tan2A + sin2A
Hence proved.
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