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.

AC

^{2 }= AB^{2}+BC^{2}AC

^{2 }= (8k)^{2}+(7k)^{2}AC

^{2 }= 64k^{2}+49k^{2}AC

^{2 }= 113k^{2}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: