Question

If sec Q=5/4 , then evaluate

1-tanA/
1+tan A
​

Answer 1

Answer:

1/4

Step-by-step explanation:

As,

secA=5/4, so,

cosA= 4/5

so, sinA= 3/5

1-tanA/1+tanA = 1-SinA/1+sinA = 1/4

Answer 2

sec =

$\frac{hypotenuse}{base} = \frac{5}{4}$

then , hypotenuse = 5x & base = 4x

$(you \: know \: \frac{5x}{4x} = \frac{5}{4} )$

Now,

${h}^{2} = {p}^{2} + {b}^{2}$

so ,

here ,

${5x}^{2} = {p}^{2} + {4x}^{2} \\ 25 {x}^{2} - 16 {x}^{2} = {p}^{2} \\ 3x= \sqrt{9 {x}^{2} } = p$

so perpendicular = 3x

Now, tan =

$\frac{perpendicular}{base} = \frac{3x}{4x} = \frac{3}{4} \\ tan \: = \frac{3}{4}$

now,

$\frac{1 - \frac{3}{4} }{1 + \frac{3}{4} } = \frac{ \frac{1}{4} }{ \frac{7}{4} } = \frac{4 \times 1}{7 \times 4} = \frac{1}{7} or \: 0.14(approx.)$

hope it's correct