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Question Number 116630 by bobhans last updated on 05/Oct/20
Given cosec x + cot x = p , find the value of cosec x =?
$$\mathrm{Given}\:\mathrm{cosec}\:\mathrm{x}\:+\:\mathrm{cot}\:\mathrm{x}\:=\:\mathrm{p}\:,\:\mathrm{find}\:\mathrm{the}\: \\ $$$$\mathrm{value}\:\mathrm{of}\:\mathrm{cosec}\:\mathrm{x}\:=? \\ $$
Answered by TANMAY PANACEA last updated on 05/Oct/20
1+cot^2 x=cosec^2 x 1=(cosecx+cotx)(cosecx−cotx) cosecx+cotx=p cosecx−cotx=(1/p) 2cosecx=(p+(1/p)) cosecx=(1/2)(p+(1/p))
$$\mathrm{1}+{cot}^{\mathrm{2}} {x}={cosec}^{\mathrm{2}} {x} \\ $$$$\mathrm{1}=\left({cosecx}+{cotx}\right)\left({cosecx}−{cotx}\right) \\ $$$${cosecx}+{cotx}={p} \\ $$$${cosecx}−{cotx}=\frac{\mathrm{1}}{{p}} \\ $$$$\mathrm{2}{cosecx}=\left({p}+\frac{\mathrm{1}}{{p}}\right) \\ $$$${cosecx}=\frac{\mathrm{1}}{\mathrm{2}}\left({p}+\frac{\mathrm{1}}{{p}}\right) \\ $$
Answered by MJS_new last updated on 05/Oct/20
(1/(sin x))+((cos x)/(sin x))=p ((1+(√(1−sin^2 x)))/(sin x))=p ⇒ sin x =((2p)/(p^2 +1)) ⇒ cosec x =((p^2 +1)/(2p))
$$\frac{\mathrm{1}}{\mathrm{sin}\:{x}}+\frac{\mathrm{cos}\:{x}}{\mathrm{sin}\:{x}}={p} \\ $$$$\frac{\mathrm{1}+\sqrt{\mathrm{1}−\mathrm{sin}^{\mathrm{2}} \:{x}}}{\mathrm{sin}\:{x}}={p} \\ $$$$\Rightarrow\:\mathrm{sin}\:{x}\:=\frac{\mathrm{2}{p}}{{p}^{\mathrm{2}} +\mathrm{1}}\:\Rightarrow\:\mathrm{cosec}\:{x}\:=\frac{{p}^{\mathrm{2}} +\mathrm{1}}{\mathrm{2}{p}} \\ $$
Answered by bemath last updated on 05/Oct/20
(1/(sin x)) + ((cos x)/(sin x)) = p ⇒1+cos x = psin x ⇒(√(1−sin^2 x)) = psin x−1 ⇒1−sin^2 x = p^2 sin^2 x−2psin x+1 (1+p^2 )sin^2 x = 2psin x sin x ≠ 0 ⇒ (1+p^2 )sin x = 2p ⇒ sin x = ((2p)/(1+p^2 )) and cosec x = ((1+p^2 )/(2p))
$$\frac{\mathrm{1}}{\mathrm{sin}\:\mathrm{x}}\:+\:\frac{\mathrm{cos}\:\mathrm{x}}{\mathrm{sin}\:\mathrm{x}}\:=\:\mathrm{p}\:\Rightarrow\mathrm{1}+\mathrm{cos}\:\mathrm{x}\:=\:\mathrm{psin}\:\mathrm{x} \\ $$$$\Rightarrow\sqrt{\mathrm{1}−\mathrm{sin}\:^{\mathrm{2}} \mathrm{x}}\:=\:\mathrm{psin}\:\mathrm{x}−\mathrm{1} \\ $$$$\Rightarrow\mathrm{1}−\mathrm{sin}\:^{\mathrm{2}} \mathrm{x}\:=\:\mathrm{p}^{\mathrm{2}} \mathrm{sin}\:^{\mathrm{2}} \mathrm{x}−\mathrm{2psin}\:\mathrm{x}+\mathrm{1} \\ $$$$\left(\mathrm{1}+\mathrm{p}^{\mathrm{2}} \right)\mathrm{sin}\:^{\mathrm{2}} \mathrm{x}\:=\:\mathrm{2psin}\:\mathrm{x} \\ $$$$\mathrm{sin}\:\mathrm{x}\:\neq\:\mathrm{0}\:\Rightarrow\:\left(\mathrm{1}+\mathrm{p}^{\mathrm{2}} \right)\mathrm{sin}\:\mathrm{x}\:=\:\mathrm{2p} \\ $$$$\Rightarrow\:\mathrm{sin}\:\mathrm{x}\:=\:\frac{\mathrm{2p}}{\mathrm{1}+\mathrm{p}^{\mathrm{2}} }\:\mathrm{and}\:\mathrm{cosec}\:\mathrm{x}\:=\:\frac{\mathrm{1}+\mathrm{p}^{\mathrm{2}} }{\mathrm{2p}} \\ $$

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