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1 + tan A tan (A/2)
=$$1+\dfrac{2tan(A/2)}{1-tan^{2}(A/2)}\times tan(A/2)$$
=$$\dfrac{1-tan^{2}(A/2)+2tan^{2}(A/2)}{1-tan^{2}(A/2)}$$
=$$\dfrac{1+tan^{2}(A/2)}{1-tan^{2}(A/2)}$$
=$$sec2(A/2)=secA$$
so the answer is option C.
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