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Question
Prove the following identities:
sec2 A . cosec2 A = tan2 A + cot2 A + 2
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Solution
L.H.S. = sec2 A . cosec2 A
= `1/(cos^2A) * 1/(sin^2A)`
= `1/(cos^2A sin^2A)`
= `(sin^2A + cos^2A)/(cos^2A sin^2A)`
= `1/(cos^2A) + 1/(sin^2A)`
= sec2 A + cosec2 A
= 1 + tan2 A + 1 + cot2 A ...(∵ sec2 A = 1 + tan2 A and cosec2 A = 1 + cot2 A)
= tan2 A + cot2 A + 2 = R.H.S.
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RELATED QUESTIONS
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sin2θ + sin2(90 – θ) = ?
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.
Factorize: sin3θ + cos3θ
Hence, prove the following identity:
`(sin^3θ + cos^3θ)/(sin θ + cos θ) + sin θ cos θ = 1`
