Question

3 tanA= 4, then value of √1-sinA/1+cosA

Answer 1

tanA=4/3
$\sqrt{1 - sin} a \binom{1 + cosa}{?}$
$\sqrt{1 - 4 \div 5 \div 1 + 3 \div 5$
$\frac{1}{ \sqrt{5} } \times \frac{5}{8}$
$8(5 \sqrt{5) }$
$40 \sqrt{5}$

Answer 2

Tan A = 4/3,

As, tan A= perpendicular/Base,

By pythagoras theorem

(perpendicular)² + (Base)² =( hypotenuse)²

4² + 3²=( hypotenuse)²

25=( hypotenuse)²

Hypotenuse=5 units

then SinA= 4/5, CosA= 3/5 where A is an acute angle of right triangle .

Now, =$\frac{\sqrt{1 - sinA}}{{1+CosA}} =\frac{\sqrt{1-\frac{4}{5}} }{1+ \frac{3}{5}}=\frac{\frac{1}{\sqrt5}} {\frac{8}{5}}=\frac{5}{8\sqrt5}=\frac{\sqrt5}{8}$