Trignonometric Ratio of Some Specific Angle
Example : In a right triangle ABC, right-angled at AB, if tan A=1, then verify that 2 sin A Cos A = 1.
Solution : In Δ ABC, tan A = BC/AB =1.
BC= AB
i.e., Let AB= BC= k, where k is a positive number.
Now, AC = √AB²+BC²
= √(k)²+(k)² = k√2
Therefore , sin A = BC/AC = 1/√2 and
cos A = AB/AC= 1/√2
So, 2 sin A cos A = 2(1/√2)(1/√2) =1,
which is the required value.