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Question
Prove that cot θ. tan (90° - θ) - sec (90° - θ). cosec θ + 1 = 0.
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Solution
LHS = cot θ. tan (90° - θ) - sec (90° - θ). cosec θ + 1
= cot θ. cot θ - cosec θ. cosec θ + 1
= (cot2θ - cosec2θ) + 1
= - 1 + 1 = 0
= RHS
Hence proved.
RELATED QUESTIONS
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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.
