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

1 Answer

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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