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Question Number 107900 by bemath last updated on 13/Aug/20
((Σ BeMath Σ)/□) Given tan x−sec x = ϑ then sin x = ?
$$\:\:\:\:\:\:\:\frac{\Sigma\:\mathcal{B}{e}\mathcal{M}{ath}\:\Sigma}{\Box} \\ $$$$\:\:{Given}\:\mathrm{tan}\:{x}−\mathrm{sec}\:{x}\:=\:\vartheta\: \\ $$$$\:{then}\:\mathrm{sin}\:{x}\:=\:? \\ $$
Answered by bemath last updated on 13/Aug/20
((sin x−1)/(cos x)) = υ ⇒ sin x−1=υ cos x ...(1) sin x +1 = −(1/υ) cos x ...(2) substitute eq (2) to eq(1) sin x+1 = −(1/υ) (((sin x−1)/υ)) υ^2 sin x + υ^2 = −sin x +1 (υ^2 +1)sin x = 1−υ^2 ⇒ sin x = ((1−υ^2 )/(1+υ^2 ))
$$\frac{\mathrm{sin}\:{x}−\mathrm{1}}{\mathrm{cos}\:{x}}\:=\:\upsilon\:\Rightarrow\:\mathrm{sin}\:{x}−\mathrm{1}=\upsilon\:\mathrm{cos}\:{x}\:…\left(\mathrm{1}\right) \\ $$$$\mathrm{sin}\:{x}\:+\mathrm{1}\:=\:−\frac{\mathrm{1}}{\upsilon}\:\mathrm{cos}\:{x}\:…\left(\mathrm{2}\right) \\ $$$${substitute}\:{eq}\:\left(\mathrm{2}\right)\:{to}\:{eq}\left(\mathrm{1}\right) \\ $$$$\mathrm{sin}\:{x}+\mathrm{1}\:=\:−\frac{\mathrm{1}}{\upsilon}\:\left(\frac{\mathrm{sin}\:{x}−\mathrm{1}}{\upsilon}\right) \\ $$$$\upsilon^{\mathrm{2}} \:\mathrm{sin}\:{x}\:+\:\upsilon^{\mathrm{2}} \:=\:−\mathrm{sin}\:{x}\:+\mathrm{1}\: \\ $$$$\left(\upsilon^{\mathrm{2}} +\mathrm{1}\right)\mathrm{sin}\:{x}\:=\:\mathrm{1}−\upsilon^{\mathrm{2}} \: \\ $$$$\Rightarrow\:\mathrm{sin}\:{x}\:=\:\frac{\mathrm{1}−\upsilon^{\mathrm{2}} }{\mathrm{1}+\upsilon^{\mathrm{2}} } \\ $$
Answered by bobhans last updated on 13/Aug/20
((∧Bobhans∧)/□) tan x−sec x ×(((tan x+sec x)/(tan x+sec x)))= υ ((tan^2 x−sec^2 x)/(tan x+sec x)) = υ →((tan^2 x−(1+tan^2 x))/(tan x+sec c))=υ tan x+sec x = −(1/υ) ⇒ ((sin x+1)/(cos x)) = −(1/υ) sin x+1 = −(1/υ)cos x
$$\:\:\:\:\:\frac{\wedge\mathcal{B}\mathrm{obhans}\wedge}{\Box} \\ $$$$\mathrm{tan}\:\mathrm{x}−\mathrm{sec}\:\mathrm{x}\:×\left(\frac{\mathrm{tan}\:\mathrm{x}+\mathrm{sec}\:\mathrm{x}}{\mathrm{tan}\:\mathrm{x}+\mathrm{sec}\:\mathrm{x}}\right)=\:\upsilon \\ $$$$\frac{\mathrm{tan}\:^{\mathrm{2}} \mathrm{x}−\mathrm{sec}\:^{\mathrm{2}} \mathrm{x}}{\mathrm{tan}\:\mathrm{x}+\mathrm{sec}\:\mathrm{x}}\:=\:\upsilon\:\rightarrow\frac{\mathrm{tan}\:^{\mathrm{2}} \mathrm{x}−\left(\mathrm{1}+\mathrm{tan}\:^{\mathrm{2}} \mathrm{x}\right)}{\mathrm{tan}\:\mathrm{x}+\mathrm{sec}\:\mathrm{c}}=\upsilon \\ $$$$\mathrm{tan}\:\mathrm{x}+\mathrm{sec}\:\mathrm{x}\:=\:−\frac{\mathrm{1}}{\upsilon}\:\Rightarrow\:\frac{\mathrm{sin}\:\mathrm{x}+\mathrm{1}}{\mathrm{cos}\:\mathrm{x}}\:=\:−\frac{\mathrm{1}}{\upsilon} \\ $$$$\mathrm{sin}\:\mathrm{x}+\mathrm{1}\:=\:−\frac{\mathrm{1}}{\upsilon}\mathrm{cos}\:\mathrm{x} \\ $$

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