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Geometry Level 3

sec ⁡ 2 ( tan ⁡ − 1 2 ) + csc ⁡ 2 ( cot ⁡ − 1 3 ) + csc ⁡ ( 2 cot ⁡ − 1 2 + cos ⁡ − 1 3 5 ) \sec^{2}\left(\tan^{-1}2\right)+\csc^{2}\left(\cot^{-1}3\right)+\csc\left(2\cot^{-1}2+\cos^{-1} \frac{3}{5}\right)

Find the value of the expression above to 3 decimal places.


The answer is 16.042.

Watsapol Kerdlarppol by Watsapol Kerdlarppol , 22, Thailand

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1 solution

Chew-Seong Cheong
Oct 23, 2016

X = sec ⁡ 2 ( tan ⁡ − 1 2 ) + csc ⁡ 2 ( cot ⁡ − 1 3 ) + csc ⁡ ( 2 cot ⁡ − 1 2 + cos ⁡ − 1 3 5 ) = 1 + tan ⁡ 2 ( tan ⁡ − 1 2 ) + 1 + cot ⁡ 2 ( cot ⁡ − 1 3 ) + csc ⁡ ( 2 tan ⁡ − 1 1 2 + tan ⁡ − 1 4 3 ) = 1 + 2 2 + 1 + 3 2 + csc ⁡ ( tan ⁡ − 1 2 ⋅ 1 2 1 − 1 2 ⋅ 1 2 + tan ⁡ − 1 4 3 ) = 5 + 10 + csc ⁡ ( tan ⁡ − 1 4 3 + tan ⁡ − 1 4 3 ) = 15 + csc ⁡ ( tan ⁡ − 1 2 ⋅ 4 3 1 − ( 4 3 ) 2 ) = 15 + csc ⁡ ( tan ⁡ − 1 24 − 7 ) = 15 + csc ⁡ ( csc ⁡ − 1 25 24 ) = 15 + 25 24 ≈ 16.042 \begin{aligned} X & = {\color{#3D99F6}\sec^2 \left(\tan^{-1} 2\right)} + {\color{#D61F06}\csc^2 \left(\cot^{-1} 3\right)} + \csc \left(2\cot^{-1} 2 + \cos^{-1} \frac 35 \right) \\ & = {\color{#3D99F6}1+\tan^2 \left(\tan^{-1} 2\right)} + {\color{#D61F06}1+\cot^2 \left(\cot^{-1} 3\right)} + \csc \left(2\tan^{-1} \frac 12 + \tan^{-1} \frac 43 \right) \\ & = {\color{#3D99F6}1+2^2} + {\color{#D61F06}1+3^2} + \csc \left(\tan^{-1} \frac {2\cdot \frac 12}{1-\frac 12 \cdot \frac 12} + \tan^{-1} \frac 43 \right) \\ & = {\color{#3D99F6}5} + {\color{#D61F06}10} + \csc \left(\tan^{-1} \frac 43 + \tan^{-1} \frac 43 \right) \\ & = 15 + \csc \left(\tan^{-1} \frac {2 \cdot \frac 43}{1-\left(\frac 43\right)^2} \right) \\ & = 15 + \csc \left(\tan^{-1} \frac {24}{-7} \right) \\ & = 15 + \csc \left(\csc^{-1} \frac {25}{24} \right) \\ & = 15 + \frac {25}{24} \approx \boxed{16.042} \end{aligned}

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Everything is clear but how come arctan(-x) =-arctan(x)?

Vighnesh Raut - 4 years, 7 months ago

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My bad, I meant how come arctan(-x) =arctan(x)

Vighnesh Raut - 4 years, 7 months ago

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Which line is it? Do you mean tan ⁡ − 1 24 − 7 = csc ⁡ − 1 25 24 \tan^{-1} \frac {24}{-7} = \csc^{-1} \frac {25}{24} ? If so, that is because the angle x x is in the second quardrant, π / 2 < x < π \pi/2 < x < \pi , where sin ⁡ x > 0 \sin x > 0 , so csc ⁡ x = 1 sin ⁡ x > 0 \csc x = \dfrac 1{\sin x} > 0 , while cos ⁡ x < 0 \cos x < 0 , therefore, tan ⁡ x < 0 \tan x < 0 .

Chew-Seong Cheong - 4 years, 7 months ago

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