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Question Number 21622 by Joel577 last updated on 29/Sep/17

If sec x + tan x = 2012 then 2011(cosec x + cot x) is equal to (A) 2011 (B) 2012 (C) 2013 (D) ((2011)/(2013)) (E) ((2013)/(2012))

$$\mathrm{If}\:\mathrm{sec}\:{x}\:+\:\mathrm{tan}\:{x}\:=\:\mathrm{2012} \\ $$$$\mathrm{then}\:\mathrm{2011}\left(\mathrm{cosec}\:{x}\:+\:\mathrm{cot}\:{x}\right)\:\mathrm{is}\:\mathrm{equal}\:\mathrm{to} \\ $$$$\left({A}\right)\:\mathrm{2011} \\ $$$$\left({B}\right)\:\mathrm{2012} \\ $$$$\left({C}\right)\:\mathrm{2013} \\ $$$$\left({D}\right)\:\frac{\mathrm{2011}}{\mathrm{2013}} \\ $$$$\left({E}\right)\:\frac{\mathrm{2013}}{\mathrm{2012}} \\ $$

Answered by abwayh last updated on 29/Sep/17

it just try; sec x+tan x=2012 (sec x−1)+tan x=2011 ; sec x+tan x+1=2013 2tan x=2011 ......(i) ; 2sec x=2013 .....(ii) 2011(cosec x+cot x)= =2tan x((1/(sin x)) +((cos x)/(sin x)) ) =2((1/(cos x)) +1) =2sec x+2 =2013+2 =2015

$$\mathrm{it}\:\mathrm{just}\:\mathrm{try}; \\ $$$$\mathrm{sec}\:\mathrm{x}+\mathrm{tan}\:\mathrm{x}=\mathrm{2012} \\ $$$$\left(\mathrm{sec}\:\mathrm{x}−\mathrm{1}\right)+\mathrm{tan}\:\mathrm{x}=\mathrm{2011}\:\:;\:\:\mathrm{sec}\:\mathrm{x}+\mathrm{tan}\:\mathrm{x}+\mathrm{1}=\mathrm{2013} \\ $$$$\mathrm{2tan}\:\mathrm{x}=\mathrm{2011}\:......\left(\mathrm{i}\right)\:\:;\:\:\mathrm{2sec}\:\mathrm{x}=\mathrm{2013}\:.....\left(\mathrm{ii}\right) \\ $$$$\mathrm{2011}\left(\mathrm{cosec}\:\mathrm{x}+\mathrm{cot}\:\mathrm{x}\right)= \\ $$$$=\mathrm{2tan}\:\mathrm{x}\left(\frac{\mathrm{1}}{\mathrm{sin}\:\mathrm{x}}\:+\frac{\mathrm{cos}\:\mathrm{x}}{\mathrm{sin}\:\mathrm{x}}\:\right) \\ $$$$=\mathrm{2}\left(\frac{\mathrm{1}}{\mathrm{cos}\:\mathrm{x}}\:+\mathrm{1}\right) \\ $$$$=\mathrm{2sec}\:\mathrm{x}+\mathrm{2} \\ $$$$=\mathrm{2013}+\mathrm{2} \\ $$$$=\mathrm{2015} \\ $$$$ \\ $$

Commented by $@ty@m last updated on 29/Sep/17

If u wont explain this I would never understand how (i) & (ii) are formed.

$${If}\:{u}\:{wont}\:{explain}\:{this}\:{I}\:{would} \\ $$$${never}\:{understand}\:{how}\:\left({i}\right) \\ $$$$\&\:\left({ii}\right)\:{are}\:{formed}. \\ $$

Commented by Tinkutara last updated on 29/Sep/17

Another way: Since sec^2 x−tan^2 x=1 ⇒ (sec x−tan x)(sec x+tan x)=1 So if sec x+tan x=2012, then sec x−tan x=(1/(2012)) Adding and subtracting we can get sec x and tan x and thus all trigonometric ratios.

$${Another}\:{way}: \\ $$$${Since}\:\mathrm{sec}^{\mathrm{2}} \:{x}−\mathrm{tan}^{\mathrm{2}} \:{x}=\mathrm{1} \\ $$$$\Rightarrow\:\left(\mathrm{sec}\:{x}−\mathrm{tan}\:{x}\right)\left(\mathrm{sec}\:{x}+\mathrm{tan}\:{x}\right)=\mathrm{1} \\ $$$${So}\:{if}\:\mathrm{sec}\:{x}+\mathrm{tan}\:{x}=\mathrm{2012},\:{then} \\ $$$$\mathrm{sec}\:{x}−\mathrm{tan}\:{x}=\frac{\mathrm{1}}{\mathrm{2012}} \\ $$$${Adding}\:{and}\:{subtracting}\:{we}\:{can}\:{get} \\ $$$$\mathrm{sec}\:{x}\:{and}\:\mathrm{tan}\:{x}\:{and}\:{thus}\:{all} \\ $$$${trigonometric}\:{ratios}. \\ $$

Commented by Joel577 last updated on 29/Sep/17

sec x − 1 ≠ tan x tan x + 1 ≠ sec x

$$\mathrm{sec}\:{x}\:−\:\mathrm{1}\:\neq\:\mathrm{tan}\:{x} \\ $$$$\mathrm{tan}\:{x}\:+\:\mathrm{1}\:\neq\:\mathrm{sec}\:{x} \\ $$

Commented by Joel577 last updated on 30/Sep/17

Thanks for the idea

$${Thanks}\:{for}\:{the}\:{idea} \\ $$

Answered by $@ty@m last updated on 29/Sep/17

sec x + tan x = 2012 (1/(cosx ))+((sinx )/(cosx ))=2012 ((1+sinx )/(cosx ))=2012 (((cos(x/2) +sin(x/2) )^2 )/(cos^2 (x/2)−sin^2 (x/2) ))=2012 ((cos(x/2) +sin(x/2))/(cos(x/2)−sin(x/2) ))=2012 ((1+tan(x/2))/(1−tan (x/2)))=2012 1+tan(x/2)=2012(1−tan (x/2)) 2013tan(x/2) =2011 2011cot(x/2)=2013 2011((cos(x/2) )/(sin(x/2) ))=2013 2011((2cos(x/2)cos(x/2) )/(2sin(x/2)cos(x/2) ))=2013 2011((2cos^2 (x/2) )/(sinx ))=2013 2011((1+cosx )/(sinx ))=2013 2011(cosecx+cotx)=2013

$$\mathrm{sec}\:{x}\:+\:\mathrm{tan}\:{x}\:=\:\mathrm{2012} \\ $$$$\frac{\mathrm{1}}{\mathrm{cos}{x}\:}+\frac{\mathrm{sin}{x}\:}{\mathrm{cos}{x}\:}=\mathrm{2012} \\ $$$$\frac{\mathrm{1}+\mathrm{sin}{x}\:}{\mathrm{cos}{x}\:}=\mathrm{2012} \\ $$$$\frac{\left(\mathrm{cos}\frac{{x}}{\mathrm{2}}\:+\mathrm{sin}\frac{{x}}{\mathrm{2}}\:\right)^{\mathrm{2}} }{\mathrm{cos}^{\mathrm{2}} \frac{{x}}{\mathrm{2}}−\mathrm{sin}^{\mathrm{2}} \frac{{x}}{\mathrm{2}}\:\:}=\mathrm{2012} \\ $$$$\frac{\mathrm{cos}\frac{{x}}{\mathrm{2}}\:+\mathrm{sin}\frac{{x}}{\mathrm{2}}}{\mathrm{cos}\frac{{x}}{\mathrm{2}}−\mathrm{sin}\frac{{x}}{\mathrm{2}}\:\:}=\mathrm{2012} \\ $$$$\frac{\mathrm{1}+\mathrm{tan}\frac{{x}}{\mathrm{2}}}{\mathrm{1}−\mathrm{tan}\:\frac{{x}}{\mathrm{2}}}=\mathrm{2012} \\ $$$$\mathrm{1}+\mathrm{tan}\frac{{x}}{\mathrm{2}}=\mathrm{2012}\left(\mathrm{1}−\mathrm{tan}\:\frac{{x}}{\mathrm{2}}\right) \\ $$$$\mathrm{2013tan}\frac{{x}}{\mathrm{2}}\:=\mathrm{2011} \\ $$$$\mathrm{2011cot}\frac{{x}}{\mathrm{2}}=\mathrm{2013}\: \\ $$$$\mathrm{2011}\frac{\mathrm{cos}\frac{{x}}{\mathrm{2}}\:}{\mathrm{sin}\frac{{x}}{\mathrm{2}}\:}=\mathrm{2013} \\ $$$$\mathrm{2011}\frac{\mathrm{2cos}\frac{{x}}{\mathrm{2}}\mathrm{cos}\frac{{x}}{\mathrm{2}}\:}{\mathrm{2sin}\frac{{x}}{\mathrm{2}}\mathrm{cos}\frac{{x}}{\mathrm{2}}\:}=\mathrm{2013} \\ $$$$\mathrm{2011}\frac{\mathrm{2cos}^{\mathrm{2}} \frac{{x}}{\mathrm{2}}\:}{\mathrm{sin}{x}\:}=\mathrm{2013} \\ $$$$\mathrm{2011}\frac{\mathrm{1}+\mathrm{cos}{x}\:\:}{\mathrm{sin}{x}\:}=\mathrm{2013} \\ $$$$\mathrm{2011}\left(\mathrm{cosec}{x}+\mathrm{cot}{x}\right)=\mathrm{2013}\:\: \\ $$$$ \\ $$$$ \\ $$

Commented by Joel577 last updated on 30/Sep/17

Thank you very much. You made it very simple

$${Thank}\:{you}\:{very}\:{much}.\:{You}\:{made}\:{it}\:{very}\:{simple} \\ $$

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