If 5 sin θ = 3, find the value of sec θ - tan θsec θ + tan θ\dfrac{\text{sec θ - tan θ}}{\text{sec θ + tan θ}}sec θ + tan θsec θ - tan θ.
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Given,
5 sin θ = 3
⇒ sin θ = 35\dfrac{3}{5}53.
Solving,
⇒sec θ - tan θsec θ + tan θ⇒1cos θ−sin θcos θ1cos θ+sin θcos θ⇒1 - sin θcos θ1 + sin θcos θ⇒1 - sin θ1 + sin θ⇒1−351+35⇒5−355+35⇒2585⇒28⇒14.\Rightarrow \dfrac{\text{sec θ - tan θ}}{\text{sec θ + tan θ}} \\[1em] \Rightarrow \dfrac{\dfrac{1}{\text{cos θ}} - \dfrac{\text{sin θ}}{\text{cos θ}}}{\dfrac{1}{\text{cos θ}} + \dfrac{\text{sin θ}}{\text{cos θ}}} \\[1em] \Rightarrow \dfrac{\dfrac{\text{1 - sin θ}}{\text{cos θ}}}{\dfrac{\text{1 + sin θ}}{\text{cos θ}}} \\[1em] \Rightarrow \dfrac{\text{1 - sin θ}}{\text{1 + sin θ}} \\[1em] \Rightarrow \dfrac{1 - \dfrac{3}{5}}{1 + \dfrac{3}{5}} \\[1em] \Rightarrow \dfrac{\dfrac{5 - 3}{5}}{\dfrac{5 + 3}{5}} \\[1em] \Rightarrow \dfrac{\dfrac{2}{5}}{\dfrac{8}{5}} \\[1em] \Rightarrow \dfrac{2}{8} \\[1em] \Rightarrow \dfrac{1}{4}.⇒sec θ + tan θsec θ - tan θ⇒cos θ1+cos θsin θcos θ1−cos θsin θ⇒cos θ1 + sin θcos θ1 - sin θ⇒1 + sin θ1 - sin θ⇒1+531−53⇒55+355−3⇒5852⇒82⇒41.
Hence, sec θ - tan θsec θ + tan θ=14\dfrac{\text{sec θ - tan θ}}{\text{sec θ + tan θ}} = \dfrac{1}{4}sec θ + tan θsec θ - tan θ=41.
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Prove the following :
cos θ tan θ = sin θ
