Question

If
cot A=1/2
Find
cot A-secA​

Answer 1

Let us consider a triangle ABC in which angle B= 90 degree and angle A= theta degree.

cot A = 1/2 = AB/BC

Let AB=k and BC=2k.

According to Pythagoras theorem,

AC2 = AB2 + BC2 = k2 + 2k2 = k2 + 4k2 = 5k2

AC = $\sqrt5$k

Now, sec A= AB/AC = k/ $\sqrt5$k = 1/$\sqrt5$

cot A- sec A= 1/2-1/$\sqrt5$ = $\sqrt5$-2/ 2$\sqrt5$

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