Question-87027

Question Number 87027 by john santu last updated on 02/Apr/20

Answered by som(math1967) last updated on 05/Apr/20

∠ D C O = a l t ∠ O A P = θ (l e t)

∠ C O Q = ∠ O A P = θ

a g a i n ∠ A D O = ∠ D C O = θ n s

[△ A D O \sim △ D C O]

O C = n s e c θ, O A = O P c o s e c θ = c o s e c θ

[∵ n = O Q]

∴ O C + O A = A C

n s e c θ + c o s e c θ = d \dots \dots . 1)

n o w O D = O C t a n θ [f r o m △ O D C]

O D = n s e c θ t a n θ

f r o m △ O D A O D = O A c o t θ

∴ O D = c o s e c θ c o t θ

n s e c θ t a n θ = c o s e c θ c o t θ

{n t a n}^{2} θ = c o t θ

{t a n}^{3} θ = \frac{1}{n} t a n θ = {(\frac{1}{n})}^{\frac{1}{3}}

∴ s e c θ = \frac{{(1 + n^{\frac{2}{3}})}^{\frac{1}{2}}}{n^{\frac{1}{3}}}

c o s e c θ = \frac{{(1 + n^{\frac{2}{3}})}^{\frac{1}{2}}}{1}

f r o m 1) n s e c θ + c o s e c θ = d

n . \frac{{(1 + n^{\frac{2}{3}})}^{\frac{1}{2}}}{n^{\frac{1}{3}}} + {(1 + n^{\frac{2}{3}})}^{\frac{1}{2}} = d

n^{\frac{2}{3}} {(1 + n^{\frac{2}{3}})}^{\frac{1}{2}} + {(1 + n^{\frac{2}{3}})}^{\frac{1}{2}} = d

{(1 + n^{\frac{2}{3}})}^{\frac{1}{2}} (1 + n^{\frac{2}{3}}) = d

{(1 + n^{\frac{2}{3}})}^{\frac{3}{2}} = d

∴ 1 + n^{\frac{2}{3}} = d^{\frac{2}{3}}