Question

$\huge \tt \pink{Question - }$
Given 15 cot A = 8, find sin A and sec A.

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Answer 1

Answer:

$\displaystyle{\boxed{\red{\sf\:\sin\:A\:=\:\dfrac{15}{17}}}}$

$\displaystyle{\boxed{\red{\sf\:\sec\:A\:=\:\dfrac{17}{8}}}}$

Step-by-step-explanation:

We have given that,

$\displaystyle{\sf\:15\:\cot\:A\:=\:8}$

We have to find $\displaystyle{\sf\:\sin\:A\:\&\:\sec\:A}$

Now,

$\displaystyle{\sf\:15\:\cot\:A\:=\:8}$

$\displaystyle{\implies\sf\:\cot\:A\:=\:\dfrac{8}{15}}$

$\displaystyle{\implies\sf\:\sqrt{cosec^2\:A\:-\:1}\:=\:\dfrac{8}{15}\:\quad\:-\:-\:-\:[\:\because\:cosec^2\:A\:=\:1\:+\:\cot^2\:A\:]}$

$\displaystyle{\implies\sf\:cosec^2\:A\:-\:1\:=\:\left(\:\dfrac{8}{15}\:\right)^2\:\quad\:-\:-\:-\:[\:Squaring\:both\:sides\:]}$

$\displaystyle{\implies\sf\:cosec^2\:A\:-\:1\:=\:\dfrac{8^2}{15^2}}$

$\displaystyle{\implies\sf\:cosec^2\:A\:-\:1\:=\:\dfrac{64}{225}}$

$\displaystyle{\implies\sf\:cosec^2\:A\:=\:\dfrac{64}{225}\:+\:1}$

$\displaystyle{\implies\sf\:cosec^2\:A\:=\:\dfrac{64\:+\:225}{225}}$

$\displaystyle{\implies\sf\:cosec^2\:A\:=\:\dfrac{289}{225}}$

$\displaystyle{\implies\sf\:cosec\:A\:=\:\sqrt{\dfrac{289}{225}}\:\quad\:-\:-\:-\:[\:Taking\:square\:roots\:]}$

$\displaystyle{\implies\sf\:\dfrac{1}{\sin\:A}\:=\:\sqrt{\dfrac{17\:\times\:17}{15\:\times\:15}}\:\quad\:-\:-\:-\:\left[\:\because\:cosec\:A\:=\:\dfrac{1}{\sin\:A}\:\right]}$

$\displaystyle{\implies\sf\:\dfrac{1}{\sin\:A}\:=\:\dfrac{17}{15}}$

$\displaystyle{\implies\underline{\boxed{\red{\sf\:\sin\:A\:=\:\dfrac{15}{17}}}}}$

Now,

$\displaystyle{\pink{\sf\:\sec\:A\:=\:\dfrac{1}{\cos\:A}}\sf\:\quad\:-\:-\:-\:[\:Trigonometric\:identity\:]}$

$\displaystyle{\implies\sf\:\sec\:A\:=\:\dfrac{1}{\sqrt{1\:-\:\sin^2\:A}}\:\quad\:-\:-\:-\:[\:\because\:\sin^2\:A\:+\:\cos^2\:A\:=\:1\:]}$

$\displaystyle{\implies\sf\:\sec\:A\:=\:\dfrac{1}{\sqrt{1\:-\:\left(\:\dfrac{15}{17}\:\right)^2}}}$

$\displaystyle{\implies\sf\:\sec\:A\:=\:\dfrac{1}{\sqrt{1\:-\:\dfrac{225}{289}}}}$

$\displaystyle{\implies\sf\:\sec\:A\:=\:\dfrac{1}{\sqrt{\dfrac{289\:-\:225}{289}}}}$

$\displaystyle{\implies\sf\:\sec\:A\:=\:\dfrac{1}{\sqrt{\dfrac{64}{289}}}}$

$\displaystyle{\implies\sf\:\sec\:A\:=\:\dfrac{1}{\dfrac{8}{17}}}$

$\displaystyle{\implies\underline{\boxed{\red{\sf\:\sec\:A\:=\:\dfrac{17}{8}}}}}$