Question

please answer this question​

Answer 1

Given→

cosA , tanA and secA

To find→

Given trignometric ratio in terms of sinA

Solution→

$→ {sin}^{2} A + {cos}^{2} A = 1 \: \: \: \\ \\ → {cos}^{2} A = 1 - {sin}^{2} A \: \: \: \\ \\ → cosA = \sqrt{1 - {sin}^{2} A}$$→ tanA = \frac{sinA}{cosA} \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ → tanA = \frac{sinA}{ \sqrt{1 - {sin}^{2}A } }$$→ secA = \frac{1}{cosA} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\→ secA = \frac{1}{ \sqrt{1 - {sin}^{2}A } }$