Find this important question of introduction to trigonometry of class 10 ncert . Sir please help me to solve the exercise 8.1 question number 7(2), its very hard to solve . If cot θ = 7/8, evaluate :(ii) cot2 θ .
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Let us assume a △ABC in which ∠B = 90° and ∠C = θ
Given:
cot θ = BC/AB = 7/8
Let BC = 7k and AB = 8k, where k is a positive real number
According to Pythagoras theorem in △ABC we get.
AC2 = AB2+BC2
AC2 = (8k)2+(7k)2
AC2 = 64k2+49k2
AC2 = 113k2
AC = √113 k
According to the sine and cos function ratios, it is written as
sin θ = AB/AC = Opposite Side/Hypotenuse = 8k/√113 k = 8/√113 and
cos θ = Adjacent Side/Hypotenuse = BC/AC = 7k/√113 k = 7/√113
Now apply the values of sin function and cos function: