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If cot θ = 7/8, evaluate :(ii) cot2 θ .Q7(2)

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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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  1. 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:

    Ncert solutions class 10 chapter 8-2

     

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