Question

if secα=5/4,then verify the value of tanα/1+tan²α=sinα/secα​

Answer 1

Step-by-step explanation:

$\sec(a) = \frac{5}{4} \\ \frac{1}{ \cos(a) } = \frac{5}{4} \\ \cos(a) = \frac{4}{5} \\ we \: know \: that \: \sin{}^{2} a + \cos {}^{2} a = 1 \\ \sin(a) = \sqrt{1 - \cos {}^{2} } a \\ \sin(a) = \sqrt{1 - ( \frac{4}{5} } ) {}^{2} \\ \sin(a) = \sqrt{1 - \frac{16}{25} } \\ \sin(a) = \sqrt{ \frac{25 - 16}{25} } \\ \sin(a) = \sqrt{ \frac{9}{25} } \\ \sin(a) = \frac{3}{5} \\ \tan(a) = \frac{ \sin(a) }{ \cos(a) } \\ \tan(a) = \frac{ \frac{3}{5} }{\frac{4}{5} } \\ \tan(a) = \frac{3}{4} \\ 1 + \tan {}^{2} a = 1 + \frac{9}{16} \\ \: \: \: \: = \frac{25}{16 } \\ \frac{ \tan(a) }{1 + \tan {}^{2} a} = \frac{ \frac{3}{4} }{ \frac{25}{16} } \\ \: \: \: \: \: \: \: = \frac{3 \times 16}{4 \times 25 } \\ \: \: \: = \frac{12}{25} \\ \\ \frac{ \sin(a) }{ \sec(a) } = \frac{ \frac{3}{5} }{ \frac{5}{4} } \\ \: \: \: \: \: = \frac{12}{25} \: \: (proved)$