In a right $Δ A B C$ , right-angled at B, if tan A = 1 then, find the value of 2 sin A cos A.

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Solution

The correct option is A 1
In right-angled $△ A B C t a n A = 1 \frac{B C}{A B} = 1 B C = A B$

Hence, ABC is an isosceles right triangle.
Let BC = AB = k

According to Pythagoras theorem,

$(Hypotenuse)^{2} = (Base)^{2} + (Altitude)^{2}$
$A C^{2} = A B^{2} + B C^{2} A C^{2} = k^{2} + k^{2} = 2 k^{2} A C = \sqrt{2 k^{2}} = k \sqrt{2}$

$s i n A = \frac{O p p o s i t e s i d e}{H y p o t e n u s e s i d e} = \frac{B C}{A C} = \frac{k}{k \sqrt{2}} = \frac{1}{\sqrt{2}}$

$c o s A = \frac{A d j a c e n t s i d e}{H y p o t e n u s e s i d e} = \frac{A B}{A C} = \frac{k}{k \sqrt{2}} = \frac{1}{\sqrt{2}}$

Then, $2 s i n A c o s A = 2 \times \frac{1}{\sqrt{2}} \times \frac{1}{\sqrt{2}} = 2 \times \frac{1}{2} = 1$

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