Question

in a triangle abc right angled at b, AB :AC=1:2. Then the value of cot A+tan C/sin C+cosB is:

Answer 1

Answer:

please refer to the above attachment for your answer

Answer 2

Answer:

The value of the given expression is

$\frac{cotA+tanC}{sinC+cosB}=\frac{4}{\sqrt{3}}$

Step-by-step explanation:

Given that in ΔABC,

$B=90^{0}\\AB:AC=1:2$ and

We are required to find the value of

$\frac{cotA+tanC}{sinC+cosB}$

In ΔABC,

Let $AB=k= > AC=2k$

From the Pythagorean theorem we have

$AB^2+BC^2=AB^2\\= > k^2+BC^2=4k^2\\= > BC=\sqrt{3}k \:units$

Therefore the required trigonometric ratios are

$cotA=\frac{AB}{BC} =\frac{k}{\sqrt{3} k} =\frac{1}{\sqrt{3}}\\tanC=\frac{AB}{BC} =\frac{k}{\sqrt{3} k} =\frac{1}{\sqrt{3}}\\SinC=\frac{AB}{AC}=\frac{k}{2k}=\frac{1}{2}\\cosB=cos90=0$

Substituting these values in the given expression we get

$\frac{cotA+tanC}{sinC+cosB}=\frac{\frac{1}{\sqrt{3}}+\frac{1}{\sqrt{3}}}{\frac{1}{2}+0}\\=\frac{4}{\sqrt{3}}$

Therefore,

The value of the given expression is

$\frac{cotA+tanC}{sinC+cosB}=\frac{4}{\sqrt{3}}$

#SPJ3