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# Prove the following identities, where the angles involved are acute angles for which the expressions are defined(ix) (cosec A – sin A)(sec A – cos A) = 1/(tan A+cotA) [Hint : Simplify LHS and RHS separately]. Q.5(9)

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Find the best way to solve this tricky question of ncert class 10 of trigonometry please help me to find out the easiest way of the question of trigonometry of exercise 8.4  . Prove the following identities, where the angles involved are acute angles for which the expressions are defined(ix) (cosec A – sin A)(sec A – cos A) = 1/(tan A+cotA) [Hint : Simplify LHS and RHS separately].

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1. (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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