Question

$\huge\color{cyan}\boxed{\colorbox{black}{Question :-}}$
Given :-
$15 \: cot \: A \: = 8,find \: sin \: A \: and \: sec \: A$
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Answer 1

Given, 15cotA = 8    ⇒ cotA = 8/`15

  Using,  1 + cot²θ = cosec²θ

In the question:

       1 + (8/15)² = cosec²A

       1 + 64/225 = cosec²A

       289/225  = cosec²A

       17/15 = cosecA

Hence,  sinA = 1/cosecA = 15/17

Using,  cotA = cosA/sinA

            8/15 = cosA/(15/17)

            8/15 = 17cosA/15

            8/17 = cosA

Hence, secA = 1/cosA = 17/8

Answer 2

⇝ Given :-

• $\rm15 \: cotA = 8$

⇝ To Find :-

• $\rm sinA \: \: and \: \: secA$

⇝ Solution :-

We Have,

$\rm 15 \ cotA = 8$

$:\longmapsto \rm cotA = \frac{8}{15}$

⏩ Squaring Both Sides :

$:\longmapsto \rm {cot}^{2} A = \frac{64}{225}$

We Know that,

$\large \bf \red\bigstar \: \: \orange{ \underbrace{ \underline{  {cosec}^{2}A = 1 + {cot}^{2} A }}}$

Hence,

$:\longmapsto \rm {cosec}^{2} A = 1 + \frac{64}{225}$

$:\longmapsto \rm {cosec}^{2} A = - \frac{225 + 64}{225}$

$:\longmapsto \rm {cosec}^{2} A = \frac{289}{225}$

$:\longmapsto \rm cosecA = \sqrt{ \frac{289}{225} }$

$:\longmapsto \rm cosecA = \frac{17}{15}$

We Know that,

$\large \bf \red\bigstar \: \: \orange{ \underbrace{ \underline{sin  A = \frac{1}{cosecA} }}}$

Hence,

$:\longmapsto \rm sinA = \frac{1}{17/15}$

$\purple{ \large :\longmapsto  \underline {\boxed{{\bf sinA = \frac{15}{17} } }}}$

We Know that,

$\large \bf \red\bigstar \: \: \orange{ \underbrace{ \underline{ cotA = \frac{cosA}{sinA} }}}$

Hence,

$:\longmapsto \rm \frac{8}{15} = \frac{cosA}{15/17}$

$:\longmapsto \rm \frac{8}{ \cancel{15}} = \frac{17cosA}{ \cancel{15}}$

$:\longmapsto \rm cosA = \frac{8}{17}$

We Know that,

$\large \bf \red\bigstar \: \: \orange{ \underbrace{ \underline{sec  A = \frac{1}{cosA} }}}$

Hence,

$:\longmapsto \rm secA = \frac{1}{8/17}$

$\purple{ \large :\longmapsto  \underline {\boxed{{\bf sinA = \frac{17}{8} } }}}$