Question: Simplify: \dfrac{1+a}{1-a} +\dfrac{1-a}{1+a} -\dfrac{1+a²}{1-a²} -\dfrac{1-a²}{1+a²}

Solution:
Given,

= \dfrac{1+a}{1-a} +\dfrac{1-a}{1+a} -\dfrac{1+a²}{1-a²} -\dfrac{1-a²}{1+a²}

= \dfrac{(1+a)(1+a) +(1-a)(1-a)}{(1-a)(1+a)} -\dfrac{1+a²}{1-a²} -\dfrac{1-a²}{1+a²}

= \dfrac{1 +2a +a² +1 -2a +a²}{1-a²} -\dfrac{1+a²}{1-a²} -\dfrac{1-a²}{1+a²}

= \dfrac{2a² +2}{1-a²} -\dfrac{1+a²}{1-a²} -\dfrac{1-a²}{1+a²}

= \dfrac{2a² +2 -(1+a²)}{1-a²} -\dfrac{1-a²}{1+a²}

= \dfrac{2a² +2 -1 -a²}{1-a²} - \dfrac{1-a²}{1+a²}

= \dfrac{1+a²}{1-a²} - \dfrac{1-a²}{1+a²}

= \dfrac{(1+a²)(1+a²) - (1-a²)(1-a²)}{(1-a²)(1+a²)}

= \dfrac{1 +2a² +a⁴ -(1-2a²+a⁴)}{1-a⁴}

= \dfrac{1 +2a² +a⁴ -1 +2a² -a⁴}{1-a⁴}

= \dfrac{4a²}{1-a⁴}
= Answer

Related Notes and Solutions:

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