Question

in a traingle ABC right angled at B , AB:AC=1:2 then the value of cot A+ tan C / Sin B +Cos B

Answer 1

Answer:

i hope it's help you

ᗷTS⟭⟬ ⁷⟬⟭

Answer 2

Given:

In a triangle, ABC right angled at B, AB: AC is 1:2

To Find:

The value of (cot A+ tan C) / (Sin C +Cos A)

Solutions:

Let us construct a triangle ABC  which is right-angled at B, in which AB: AC is equal to 1:2 then we can take the values as

AB=x

AC=2x

now to find the value of the BC we will use the Pythagoras theorem which will go as,

$(2x)^2=x^2+BC^2\\BC^2=4x^2-x^2\\BC=\sqrt{3x^2}\\BC=\sqrt{3}x$

So BC=$\sqrt{3} x$

Now we can say that

p=x

h=2x

b=$\sqrt{3} x$

Now putting all the values accordingly with respect to the trigonometry identity

therefore,

$=\frac{cotA+tanC}{sinC+cosA} \\=\frac{\frac{x}{\sqrt{3}x } +\frac{x}{\sqrt{3} x} }{\frac{x}{2x} +\frac{x}{2x} } \\=\frac{\frac{2}{\sqrt{3} } }{\frac{2}{2} }\\=\frac{2}{\sqrt{3} }$

Hence, the value of (cot A+ tan C) / (Sin C +Cos A) is $\frac{2}{\sqrt{3} }$.