Question

Prove that (cotA - CosA)/(cotA+ cosA)= (cosec A- 1)/(CosecA+1)

Answer 1

Answer:

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Answer 2

To prove that, $\dfrac{\cot A - \cos A}{\cot A + \cos A}=\dfrac{\csc A -1}{\csc A +1}$.

Solution:

L.H.S. = $\dfrac{\cot A - \cos A}{\cot A + \cos A}$

Using the trigonometric identity,

$\cot A$ = $\dfrac{\cos A}{\sin A}$

= $\dfrac{\dfrac{\cos A}{\sin A}- \cos A}{\dfrac{\cos A}{\sin A}+ \cos A}$

$=\dfrac{\dfrac{\cos A-\cos A\sin A}{\sin A}}{\dfrac{\cos A+\cos A\sin A}{\sin A}}$

$=\dfrac{\cos A-\cos A\sin A}{\cos A+\cos A\sin A}$

Taking $\cos A$ as common numerator and denominator, we get

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

$=\dfrac{1-\sin A}{1+\sin A}$

Dividing nemerator and denominator by $\sin A$, we get

$=\dfrac{\dfrac{1-\sin A}{\sin A}}{\dfrac{1+\sin A}{\sin A}}$

$=\dfrac{\dfrac{1}{\sin A}-1}{\dfrac{1}{\sin A}+1}$

Using the trigonometric identity,

$\csc A$ = $\dfrac{1}{\sin A}$

= $\dfrac{\csc A -1}{\csc A +1}$

= R.H.S., proved.

Hence, $\dfrac{\cot A - \cos A}{\cot A + \cos A}=\dfrac{\csc A -1}{\csc A +1}$, proved.