Trigonometry! #75

Geometry Level 3

If 7 csc ⁡ θ − 3 cot ⁡ θ = 7 7\csc\theta-3\cot\theta=7 then find the value of ∣ 7 cot ⁡ θ − 3 csc ⁡ θ ∣ \lvert7\cot\theta-3\csc\theta\rvert

This problem is part of the set Trigonometry .


The answer is 3.

Omkar Kulkarni by Omkar Kulkarni , 22, India

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4 solutions

Jerome Polin
Feb 1, 2015

I dont know if somebody has a simplier solution.

7cscx - 3 cotx= 7 We square both side

49csc^2 x + 9cot^2 x - 42cscx.cotx = 49 49(cot^2x+1) + 9 cot^2 x - 42cscx.cotx=49 58cot^2x - 42 cscx.cotx= 0 58 cot^2x=42 cscx.cotx 58 cotx= 42 csc x 29 cotx= 21 csc x

Substitute to the equation 29/3 (cotx) - 3cotx= 7 (20/3) cot x= 7 Cotx= 21/20 Cscx= 29/20

So for 7cotx-3csc x= 7.21/20 - 3.29/20 = 60/20=3

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please reply this method is correct or wrong

7cscx-3cotx=7 7/sinx-3cosx/sinx=7 (7-3cosx)/sinx=7 7-3cosx=7sinx 7(1-sinx)=3cosx 1-sinx=3/7cosx---------------------------eq 1 let 7cotx-3cscx=k thefore by same above method we get 7-3cosx=ksinx we know 1-sin^2x=cos^2 1+sinx)(1-sinx)=cos^2 1+sinx)3/7cosx=cos^2 now simplifing it we get 7-3cosx=3sinx thus k=3

TK Venkatesh - 6 years, 4 months ago

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Could you please use latex? It's hard to understand what you're trying to say.

Omkar Kulkarni - 6 years, 4 months ago
Marvin Chong
Feb 17, 2020

Look From Left To Right Look From Left To Right

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Omkar Kulkarni
Feb 25, 2015

7 csc ⁡ θ − 3 cot ⁡ θ = 7 7\csc\theta-3\cot\theta=7

Squaring both sides,

49 csc ⁡ 2 θ − 42 csc ⁡ θ cot ⁡ θ + 9 cot ⁡ 2 θ = 49 49\csc^{2}\theta-42\csc\theta\cot\theta+9\cot^{2}\theta=49

49 ( 1 + cot ⁡ 2 θ ) − 42 csc ⁡ θ cot ⁡ θ + 9 ( csc ⁡ 2 θ − 1 ) = 49 49(1+\cot^{2}\theta)-42\csc\theta\cot\theta+9(\csc^{2}\theta-1)=49

49 + 49 cot ⁡ 2 θ − 42 csc ⁡ θ cot ⁡ θ + 9 csc ⁡ 2 θ − 9 = 49 49+49\cot^{2}\theta-42\csc\theta\cot\theta+9\csc^{2}\theta-9=49

49 cot ⁡ 2 θ − 42 csc ⁡ θ cot ⁡ θ + 9 csc ⁡ 2 θ = 9 49\cot^{2}\theta-42\csc\theta\cot\theta+9\csc^{2}\theta=9

( 7 cot ⁡ θ ) 2 − 2 ( 7 cot ⁡ θ ) ( ( 3 csc ⁡ θ ) + ( 3 csc ⁡ θ ) 2 = 3 2 (7\cot\theta)^{2}-2(7\cot\theta)((3\csc\theta)+(3\csc\theta)^{2}=3^{2}

( 7 cot ⁡ θ − 3 csc ⁡ θ ) 2 = 3 2 (7\cot\theta-3\csc\theta)^{2}=3^{2}

7 cot ⁡ θ − 3 csc ⁡ θ = ± 3 7\cot\theta-3\csc\theta=\pm3

∣ 7 cot ⁡ θ − 3 csc ⁡ θ ∣ = 3 \boxed{\lvert7\cot\theta-3\csc\theta\rvert=3}

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Thanks. I updated the question to ask for the absolute value, which is 3.

Calvin Lin Staff - 6 years, 4 months ago

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OK. Thanks for your reply and your attention.

Luiz Ponce Alonso Ponce - 6 years, 4 months ago

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