Question

Find the number of 2-digit positive integers N such that tanN15° + cot 15° is an integer.​

Answer 1

Step-by-step explanation:

*Given: Find the number of 2-digit positive integers N such that tanN15° + cotN15° is an integer.

To Find: Value of N.

Solution: We know that tan45°=1 and cot 45°=1.

Thus,

One have to assume N,such that tanN15° + cotN15° will become an integer.

Let N=15

$\tan(15 \times 15°) + \cot(15 \times 15°) \\ \\ = \tan(225°) + \cot(225° ) \\ \\ = \tan(180° + 45°) + \cot(180°+ 45°)$

we know that

$tan(180° + \theta) = tan \: \theta \\ \\ cot(180° + \theta) = cot \: \theta$

Thus,

$= \tan(45)° + \cot(45°) \\ \\ = 1 + 1 \\ \\ = 2 \: \in \: Z$

Final answer:

For N=15

$\bold{\red{\tan(N15°) + \cot(N15°) = 2}}$

Hope it helps you.

Answer 2

Given : tan (N*15°)  + cot ( N*15°) is an integer

To Find : number of 2-digit positive integers N

Solution:

tan (N*15°)  + cot ( N*15°)

= Sin (N*15°)/Cos (N *15°) +  Cos (N *15°) /Sin (N*15°)

= ( Sin² (N*15°) + Cos² (N *15°) )/ Sin (N*15°)  Cos (N *15°)

=  1/ Sin (N*15°)  Cos (N *15°)

= 2/ 2 Sin (N*15°)  Cos (N *15°)

= 2/ Sin (N * 30°)

Sin (N * 30°) =±1  or    Sin (N * 30°) = ±1/2     will give an integer

Sin (N * 30°) =  Sin (180°n  + 90° )

=> N * 30° =180°n  + 90°

=> N = 6n  + 3

k  from 2  to  16 will given N = 15  to 99  Hence 15 such numbers

N = 15 , 21 ,27 , 33 , 39 , 45 , 51 , 57 ,63 , 69 ,  75 , 81 ,87 , 93 ,99

Sin (N * 30°) =±1/2  

N * 30°   = n*180° ± 30°  

=> N  =  6n ± 1

N  = 6n + 1        n = 2 to  16   N from 13 to 97  Hence 15 such numbers

N =  13 , 19,  25 , 31 ,  37 ,43 ,  49 , 55 ,61 ,  67 ,  73 , 79 , 85 , 91 , 97

N = 6n - 1           n = 2 to  16   N from 11 to 95    Hence 15 such numbers

N = 11, 17 , 23, 29 ,35 , 41 , 47 , 53 , 59,  65 , 71 , 77 , 83 , 89 , 95

Hence 15 + 15 + 15 = 45  Such 2 Digits numbers are possible

where tan (N*15°)  + cot ( N*15°) is integer  

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