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Question
Prove the following trigonometric identities.
`[tan θ + 1/cos θ]^2 + [tan θ - 1/cos θ]^2 = 2((1 + sin^2 θ)/(1 - sin^2 θ))`
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Solution 1
LHS = (tan θ + sec θ)2 + (tan θ - sec θ)2
`"LHS" = tan^2 θ + sec^2 θ + cancel(2 tan θ. sec θ) + tan^2 θ + sec^2 θ - cancel(2 tan θ. sec θ) ...{(a^2 + b^2 = a^2 + 2ab + b^2),(a^2 - b^2 = a^2 - 2ab + b^2):}`
LHS = `2 tan^2θ + 2 sec^2θ`
LHS = `2[tan^2θ + sec^2θ]`
LHS = `2[sin^2 θ/cos^2 θ + 1/cos^2 θ]`
LHS = `2((sin^2 θ + 1)/cos^2 θ)`
LHS = `2((1 + sin^2 θ)/(1 - sin^2θ)) ...{(sin^2θ + cos^2θ = 1),(∴ cos^2θ = 1 - sin^2θ):}`
RHS = `2((1 + sin^2 θ)/(1 - sin^2θ))`
LHS = RHS
Solution 2
LHS = `[tan θ + 1/cos θ]^2 + [tan θ - 1/cos θ]^2 ...{(a + b)^2 + (a - b)^2 = 2(a^2 + b^2)}`
LHS = `2[tan^2θ + 1/cos^2θ]`
LHS = `2[sin^2 θ/cos^2 θ + 1/cos^2 θ]`
LHS = `2((sin^2 θ + 1)/cos^2 θ)`
LHS = `2((1 + sin^2 θ)/(1 - sin^2θ)) ...{(sin^2θ + cos^2θ = 1),(∴ cos^2θ = 1 - sin^2θ):}`
RHS = `2((1 + sin^2 θ)/(1 - sin^2θ))`
LHS = RHS
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