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If sec(-) + tan(-) = p, …

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If sec(-) + tan(-) = p, then find th value of cosec(-).
  • 1 answers

Sia ? 6 years, 6 months ago

{tex}sec\ \theta+ tan\ \theta  = p{/tex} ...(i)
Also {tex}sec^2 \theta  - tan^2 \theta  = 1{/tex}
{tex}\Rightarrow{/tex} (sec {tex}\theta{/tex} - tan {tex}\theta{/tex}) (sec {tex}\theta{/tex} + tan {tex}\theta{/tex}) = 1
{tex}\Rightarrow{/tex} p(sec {tex}\theta{/tex} - tan {tex}\theta{/tex}) = 1
[using equation (i)]
{tex}\Rightarrow{/tex} sec {tex}\theta{/tex} - tan {tex}\theta{/tex} {tex}=\frac{1}{p}{/tex} ...(ii)
(ii) - (i) we get
{tex}-2 tan{/tex} {tex}\theta{/tex} {tex}=\frac{1-p^{2}}{p}{/tex}
{tex}\Rightarrow{/tex}- tan {tex}\theta{/tex} {tex}=\frac{1-p^{2}}{2 p}{/tex}
{tex}\Rightarrow{/tex}- cot {tex}\theta{/tex} {tex}=\frac{2 p}{1-p^{2}}{/tex}
cot {tex}\theta{/tex} {tex}=\left(\frac{2 p}{1-p^{2}}\right)^{2}{/tex}
{tex}=\frac{-4 p^{2}}{\left(1-p^{2}\right)^{2}}{/tex}
{tex}\Rightarrow{/tex} {tex}cosec^2{/tex} {tex}\theta{/tex} - 1 {tex}=\frac{-4 p^{2}}{\left(1-p^{2}\right)^{2}}{/tex}
{tex}\Rightarrow{/tex} {tex}cosec^2{/tex} {tex}\theta{/tex} {tex}=\frac{-4 p^{2}}{\left(1-p^{2}\right)^{2}}+1=\frac{-4 p^{2}+\left(1-p^{2}\right)^{2}}{\left(1-p^{2}\right)^{2}}{/tex}
{tex}cosec^2{/tex} {tex}\theta{/tex} {tex}=\frac{-4 p^{2}+1+p^{4}-2 p^{2}}{\left(1-p^{2}\right)^{2}}{/tex}
{tex}=\frac{\left(1+p^{2}\right)^{2}}{\left(1-p^{2}\right)^{2}}{/tex}
{tex}\Rightarrow{/tex} {tex}cosec\ \theta{/tex} {tex}=\frac{1+p^{2}}{1-p^{2}}{/tex}

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