AB is a vertical pole. The end A is on the level ground. ‘C’ is the middle point of AB. P is a point on the level ground. The portion CB subtends an angle βat P. If AP=nAB, then tanβ=

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nn2+1

n2n2+1

nn2−1

n2n2−1

answer is B.

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Detailed Solution

We have AB=h⇒AC=BC=h/2Also AP=n.AB=nh∴tanα=h/2nh=12n,tanα+β=hnh=1n Now tanβ=tanα+β−α=tanα+β−tanα1+tanα+βtanα=1n−12n1+1n.12n=n2n2+1

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