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Question
Prove that `(sin θ + tan θ)/(cos θ) = tan θ (1 + sec θ)`.
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Solution
L.H.S. = `(sin θ + tan θ)/(cos θ)`
= `(sin θ)/(cos θ) + (tan θ)/(cos θ)`
= tan θ + tan θ sec θ
= tan θ (1 + sec θ)
= R.H.S.
∴ `(sin θ + tan θ)/(cos θ) = tan θ (1 + sec θ)`
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tan2θ – sin2θ = tan2θ × sin2θ. For proof of this complete the activity given below.
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.
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