Question

In right triangle ABC, right angled at c, if tanA=1 then the value of 2sinA cosA is​
Steps are needed
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Answer 1

Given:

In right triangle ABC Angle C=90 degree

tan A=1

To find:

2sin Acos A=12sinAcosA=?

Solution:

tan A=$\sf \dfrac{Perpendicular\;side}{Base}=\frac{1}{1}$

⇢BC=k

⇢AC=k

Using Pythagoras theorem

$\sf{(Hypotenuse)^2=(Perpendicular)^2+(Base)^2(Hypotenuse)}$

⇢$\sf{AB^2=AC^2+BC^2AB}$

⇢$\sf{AB^2=k^2+k^2=2k^2AB}$

⇢AB=$\sqrt{2k^2}=k\sqrt{2}$

⇢Sin A=$\sf \dfrac{Perpendicular\;side}{Hypotenuse}$ =$\sf \dfrac{BC}{AB}=\dfrac{k}{k\sqrt{2}}$ =$\sf \dfrac{1}{\sqrt{2}}$

⇢cos A=$\sf \dfrac{Base}{Hypotenuse}$ =$\sf \dfrac{AC}{AB}=\sf \dfrac{k}{k\sqrt{2}}$ =$\sf \dfrac{1}{\sqrt{2}}$

Lets Substitute the values

2sinAcos A=2$\times \dfrac{1}{(\</p><p>sqr{2}}\times \dfrac{1}{(\sqrt{2}}$

⇢2sinA cos A=$\sf \dfrac{2}{(\sqr {2})^2}$ =$\sf \dfrac{2}{2}$

⇢2sinA cos A= 12sinA cosA=1