In a right triangle ABC, right angles at B, If tan = 1, then verify that 2sinAcosA = 1
Answers
Step-by-step explanation:
In right triangle ABC
Angle C=90 degree
tan A=1
To verify that
2sin Acos A=12sinAcosA=1
Solution:
tan A=1
tan A=\frac{Perpendicular\;side}{Base}=\frac{1}{1}tanA=
Base
Perpendicularside
=
1
1
BC=k
AC=k
Using Pythagoras theorem
(Hypotenuse)^2=(Perpendicular)^2+(Base)^2(Hypotenuse)
2
=(Perpendicular)
2
+(Base)
2
AB^2=AC^2+BC^2AB
2
=AC
2
+BC
2
AB^2=k^2+k^2=2k^2AB
2
=k
2
+k
2
=2k
2
AB=\sqrt{2k^2}=k\sqrt{2}AB=
2k
2
=k
2
Sin A=\frac{Perpendicular\;side}{Hypotenuse}=\frac{BC}{AB}=\frac{k}{k\sqrt{2}}=\frac{1}{\sqrt{2}}SinA=
Hypotenuse
Perpendicularside
=
AB
BC
=
k
2
k
=
2
1
cos A=\frac{Base}{Hypotenuse}=\frac{AC}{AB}=\frac{k}{k\sqrt{2}}=\frac{1}{\sqrt{2}}cosA=
Hypotenuse
Base
=
AB
AC
=
k
2
k
=
2
1
Substitute the values
2sinAcos A=2\times \frac{1}{\sqr{2}}\times \frac{1}{\sqrt{2}}2sinAcosA=2×
\sqr2
1
×
2
1
2sinA cos A=\frac{2}{(\sqrt{2})^2}=\frac{2}{2}2sinAcosA=
(
2
)
2
2
=
2
2
2sinA cos A=12sinAcosA=1
Hence, verified.
