4. Given 15 cot A = 8, find sin A and sec A.

Let us assume a right angled triangle ABC, right angled at B Given: 15 cot A = 8 So, Cot A = 8/15 We know that, cot function is the equal to the ratio of length of the adjacent side to the opposite side. Therefore, cot A = Adjacent side/Opposite side = AB/BC = 8/15 Let AB be 8k and BC will be 15k Where, k is a positive real number. According to the Pythagoras theorem, the squares of the hypotenuse side is equal to the sum of the squares of the other two sides of a right angle triangle and we get, AC2=AB2 + BC2 Substitute the value of AB and BC AC2= (8k)2 + (15k)2 AC2= 64k2 + 225k2 AC2= 289k2 Therefore, AC = 17k Now, we have to find the value of sin A and sec A We know that, Sin (A) = Opposite side /Hypotenuse Substitute the value of BC and AC and cancel the constant k in both numerator and denominator, we get Sin A = BC/AC = 15k/17k = 15/17 Therefore, sin A = 15/17 Since secant or sec function is the reciprocal of the cos function which is equal to the ratio of the length of the hypotenuse side to the adjacent side. Sec (A) = Hypotenuse/Adjacent side Substitute the Value of BC and AB and cancel the constant k in both numerator and denominator, we get, AC/AB = 17k/8k = 17/8 Therefore sec (A) = 17/8