Question

If 5 cot A = 8, then get the value of sin A and sec A​

Answer 1

Answer:

Sin A = 5/✓89

Sec A= ✓89/8

Step-by-step explanation:

for steps look at the attached image above.

Answer 2

Answer:

Step-by-step explanation:

We have,

$\tt{5\,cot(A)=8}$

$\tt{\implies\,cot(A)=\dfrac{8}{5}}$

$\tt{\implies\,\dfrac{cos(A)}{sin(A)}=\dfrac{8}{5}}$

$\tt{\implies\,cos(A)=\dfrac{8}{5}\,sin(A)}$

$\tt{\implies\,cos^2(A)=\dfrac{64}{25}\,sin^2(A)}$

$\tt{\implies\,1-sin^2(A)=\dfrac{64}{25}\,sin^2(A)}$

$\tt{\implies\,25-25\,sin^2(A)=64\,sin^2(A)}$

$\tt{\implies\,25=64\,sin^2(A)+25\,sin^2(A)}$

$\tt{\implies\,25=89\,sin^2(A)}$

$\tt{\implies\,\dfrac{25}{89}=sin^2(A)}$

$\tt{\implies\,sin(A)=\dfrac{5}{\sqrt{89}}}$

Now

$\tt{tan(A)=\dfrac{5}{8}}$

$\tt{\implies\,tan^2(A)=\dfrac{25}{64}}$

$\tt{\implies\,sec^2(A)-1=\dfrac{25}{64}}$

$\tt{\implies\,sec^2(A)=\dfrac{25}{64}+1}$

$\tt{\implies\,sec^2(A)=\dfrac{25+64}{64}}$

$\tt{\implies\,sec^2(A)=\dfrac{89}{64}}$

$\tt{\implies\,sec(A)=\dfrac{\sqrt{89}}{8}}$