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Evaluate: (cosec A – sin A)(sec A – cos A) = 1/(tan A + cotA)

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Evaluate:

(cosec A – sin A)(sec A – cos A) = 1/(tan A + cotA)

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(cosec A – sin A)(sec A – cos A) = 1/(tan A + cotA)

First, find the simplified form of L.H.S

L.H.S. = (cosec A – sin A)(sec A – cos A)

Now, substitute the inverse and equivalent trigonometric ratio forms

= (1/sin A – sin A)(1/cos A – cos A)

= [(1-sin2A)/sin A][(1-cos2A)/cos A]

= (cos2A/sin A)×(sin2A/cos A)

= cos A sin A

Now, simplify the R.H.S

R.H.S. = 1/(tan A+cotA)

= 1/(sin A/cos A +cos A/sin A)

= 1/[(sin2A+cos2A)/sin A cos A]

= cos A sin A

L.H.S. = R.H.S.

(cosec A – sin A)(sec A – cos A) = 1/(tan A+cotA)

Hence proved

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