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Question
Prove that `(cos^2theta)/(sintheta) + sintheta` = cosec θ
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Solution
L.H.S = `(cos^2theta)/(sintheta) + sintheta`
= `(cos^2theta + sin^2theta)/sintheta`
= `1/sintheta` .......[∵ sin2θ + cos2θ = 1]
= cosec θ
= R.H.S
∴ `(cos^2theta)/(sintheta) + sintheta` = cosec θ
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= `square (1 - (sin^2theta)/(tan^2theta))`
= `tan^2theta (1 - square/((sin^2theta)/(cos^2theta)))`
= `tan^2theta (1 - (sin^2theta)/1 xx (cos^2theta)/square)`
= `tan^2theta (1 - square)`
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