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

using the identity cosec2A = 1+cot2A.

Asked by Abhisek | 1 year ago |  172

##### Solution :-

With the help of identity function, cosec2A = 1+cot2A, let us prove the above equation.

L.H.S. = $$\dfrac{(cos A–sin A+1)}{(cos A+sin A–1)}$$

Divide the numerator and denominator by sin A, we get

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

We know that $$\dfrac{ cos A}{sin A}$$ = cot A

and $$\dfrac{1}{sin A}$$ = cosec A

$$\dfrac{(cot A – 1 + cosec A)}{(cot A+ 1 – cosec A)}$$

$$\dfrac{ (cot A – cosec^2A + cot^2A + cosec A)}{(cot A+ 1 – cosec A) (cosec^2A – cot^2A)}=1$$

$$\dfrac{ [(cot A + cosec A) – (cosec^2A – cot^2A)]}{(cot A+ 1 – cosec A)}$$

$$\dfrac{ [(cot A + cosec A) – (cosec A + cot A)(cosec A – cot A)]}{(1 – cosec A + cot A)}$$

=  $$\dfrac{(cot A + cosec A)(1 – cosec A + cot A)}{(1 – cosec A + cot A)}$$

=  cot A + cosec A = R.H.S.

Therefore, $$\dfrac{(cos A–sin A+1)}{(cos A+sin A–1)}$$ = cosec A + cot A

Hence Proved

Answered by Pragya Singh | 1 year ago

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