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Question
If tan θ = 2, where θ is an acute angle, find the value of cos θ.
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Solution
1 + tan2 θ = sec2 θ
∴ 1 + (2)2 = sec2 θ
∴ sec2 θ = 1 + 4
= 5
sec θ = `sqrt(5)`
cos θ = `1/(sec theta)`
∴ cos θ = `1/sqrt(5)`
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Activity:
L.H.S. = `square`
= `square (1 - (sin^2θ)/(tan^2θ))`
= `tan^2θ (1 - square/((sin^2θ)/(cos^2θ)))`
= `tan^2θ (1 - (sin^2θ)/1 xx (cos^2θ)/square)`
= `tan^2θ (1 - square)`
= `tan^2θ xx square` ...[1 – cos2θ = sin2θ]
= R.H.S.
