Question

In right triangle ABC, right-angled at B, if \[{\text{tan A = 1}}\], then verify that \[{\text{2 sin A cos A = 1}}\].

Answer 1

Hint:
As we know that according to the question it is given that the right angle triangle ABC, right-angled at B. So by using simple trigonometry functions we will calculate the value of hypotenuse in terms of other two sides and find \[{\text{sin A}}\], \[{\text{cos A}}\] and put it in the equation to find its value. Then we will be able to verify the equation.

Complete step by step solution:

In ∆ABC, it is given that \[\angle {\text{ABC}} = {90^0}\]
As we all know that \[{\text{tan A = }}\dfrac{{{\text{side opposite to angle A}}}}{{{\text{side adjacent to angle A}}}}\] and it is given that \[{\text{tan A = 1}}\]
Therefore, \[{\text{tan A = }}\dfrac{{{\text{BC}}}}{{{\text{AB}}}} = 1\]
Therefore, \[{\text{BC}} = {\text{AB}}\]
But let us suppose \[{\text{BC}} = {\text{AB = k}}\] where, k is a positive number.
Now we have to find the value of AC by using Pythagoras theorem we know that \[\Rightarrow {\text{A}}{{\text{C}}^2} = {\text{A}}{{\text{B}}^2}{\text{ + B}}{{\text{C}}^2}\]
Therefore \[{\text{AC}} = \sqrt {{\text{A}}{{\text{B}}^2}{\text{ + B}}{{\text{C}}^2}} \]
By putting the value of AB and BC in above equation, we get
\[\Rightarrow {\text{AC}} = \sqrt {{{\text{k}}^2}{\text{ + }}{{\text{k}}^2}} \]
\[\therefore {\text{AC}} = {\text{k}}\sqrt 2 \]
Now, as we all know that \[{\text{sin A = }}\dfrac{{{\text{side opposite to angle A}}}}{{{\text{hypotenuse}}}}\] and \[{\text{cos A = }}\dfrac{{{\text{side adjacent to angle A}}}}{{{\text{hypotenuse}}}}\]
Therefore we will calculate the value of \[{\text{sin A}}\]and \[{\text{cos A}}\]
\[\therefore {\text{sin A = }}\dfrac{{{\text{BC}}}}{{{\text{AC}}}} = \dfrac{{\text{k}}}{{{\text{k}}\sqrt 2 }} = \dfrac{1}{{\sqrt 2 }}\]
\[\therefore {\text{cos A = }}\dfrac{{{\text{AB}}}}{{{\text{AC}}}} = \dfrac{{\text{k}}}{{{\text{k}}\sqrt 2 }} = \dfrac{1}{{\sqrt 2 }}\]
Now we have to find the value of \[{\text{2 sin A cos A}}\]
So by putting the value of \[{\text{sin A}}\]and \[{\text{cos A}}\] in the above equation, we get
\[\Rightarrow {\text{2 sin A cos A = 2}} \times \dfrac{1}{{\sqrt 2 }} \times \dfrac{1}{{\sqrt 2 }}\]
\[\Rightarrow {\text{2 sin A cos A = }}\dfrac{2}{{\sqrt 2 \times \sqrt 2 }}\]
\[\Rightarrow {\text{2 sin A cos A = }}\dfrac{2}{{{{(\sqrt 2 )}^2}}}\]
\[\Rightarrow {\text{2 sin A cos A = }}\dfrac{2}{2}\]
\[\Rightarrow {\text{2 sin A cos A = 1}}\]

Therefore, \[{\text{2 sin A cos A = 1}}\]
Hence proved.

Note:
Right Triangle is a triangle where one of its interior angles is a right angle (90 degrees). The relation between the sides and angles of a right triangle is the basis for trigonometry. The side opposite the right angle is called the hypotenuse. The sides adjacent to the right angle are called legs.
Pythagoras theorem stated that In a right angled triangle the square of the long side is equal to the sum of the squares of the other two sides.
We have to remember all the trigonometry formulas
\[{\text{sin A = }}\dfrac{{{\text{side opposite to angle A}}}}{{{\text{hypotenuse}}}}\], \[{\text{cos A = }}\dfrac{{{\text{side adjacent to angle A}}}}{{{\text{hypotenuse}}}}\], \[{\text{tan A = }}\dfrac{{{\text{side opposite to angle A}}}}{{{\text{side adjacent to angle A}}}}\], \[{\text{cot A = }}\dfrac{1}{{{\text{tan A}}}}\], \[{\text{sec A = }}\dfrac{1}{{{\text{cos A}}}}\], \[{\text{cosec A = }}\dfrac{1}{{{\text{sin A}}}}\]