Prove the following identities, where the angles involved are acute angles for which the expressions are defined.(cosec A – sin A)(sec A – cos A) = $$\dfrac{1}{(tan A+cotA)}$$

Asked by Abhisek | 1 year ago |  163

##### Solution :-

(cosec A – sin A)(sec A – cos A)

$$\dfrac{1}{tan A+cotA}$$

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

$$( \dfrac{1}{sin A} – sin A) ( \dfrac{1}{cos A }– cos A)$$

$$( \dfrac{1-sin^2A}{sin A})( \dfrac{1-cos^2A}{cos A})$$

$$\dfrac{cos^2A}{sin A}× \dfrac{sin^2A}{cos A}$$

= cos A sin A

Now, simplify the R.H.S

R.H.S. = $$\dfrac{1}{tan A+cotA}$$

$$\dfrac{1}{ \dfrac{sin A}{cos A}+ \dfrac{cos A}{sin A}}$$

$$\dfrac{1}{sin^2A+cos^2A}$$

$$\dfrac{1}{sin A cos A}$$

= cos A sin A

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

(cosec A – sin A)(sec A – cos A)

$$\dfrac{1}{tan A+cotA}$$

Hence proved

Answered by Pragya Singh | 1 year ago

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