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Question

$$\dfrac {\cos A}{1 - \tan A} - \dfrac {\sin^{2}A}{\cos A - \sin A} = \sin A + \cos A$$.


Solution

We have,

L.H.S. = $$\dfrac{cos \, A}{1 \, - \, tan \, A} \, + \, \dfrac{sin^2 \, A}{sin \, A \, - \, cos \, A}$$

$$\Rightarrow \, \,$$ L.H.S. = $$\dfrac{cos \, A}{1 \, - \, \dfrac{sin \, A}{cos \, A}} \, + \, \dfrac{sin^2 \, A}{sin \, A \, - \, cos \, A}$$

$$\Rightarrow \, \,$$ L.H.S. = $$\dfrac{cos \, A}{\dfrac{cos \, A \, - \, sin \, A}{cos \, A}} \, + \, \dfrac{sin^2 \, A}{sin \, A \, - \, cos \, A}$$

$$\Rightarrow \, \,$$ L.H.S. = $$\dfrac{cos^2 \, A}{cos \, A \, - \, sin \, A} \, + \, \dfrac{sin^2 \, A}{sin \, A \, - \, cos \, A}$$

$$\Rightarrow \, \,$$ L.H.S. = $$\dfrac{cos^2 \, A}{cos \, A \, - \, sin \, A} \, - \, \dfrac{sin^2 \, A}{cos \, A \, - \, sin \, A}$$ 

$$\Rightarrow \, \,$$ L.H.S. = $$\dfrac{cos^2 \, A \, - \, sin^2 \, A}{cos \, A \, - \, sin \, A}$$

$$\Rightarrow \, \,$$ L.H.S. = $$\dfrac{(cos \, A \, + \, sin \, A) \, (cos \, A \, - \, sin \, A)}{cos \, A \, - \, sin \, A} \, = \, cos \, A \, + \, sin \, A \, = \, R.H.S.$$

Mathematics

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