Given sinA=cosB+cosC, this specific structure in triangle problems often leads to an equilateral or specific triangle case. If we test an equilateral triangle where A=B=C=60∘:
sin60∘=23
cos60∘+cos60∘=21+21=1
Since 23=1, it is not equilateral.
However, using tan(2B)+cot(2B), the expression is simply 2cscB. For the identity to hold generally in such problems, the value simplifies to a constant. By testing the identity A=90∘,B+C=90∘, one can show the value is 2.
Correct Option:(d)2
Explanation
1. Apply trigonometric identities:
In a triangle, A+B+C=π, so A=π−(B+C).
Thus, sinA=sin(π−(B+C))=sin(B+C).
The given equation is:
sin(B+C)=cosB+cosC
Using the sum-to-product formulas:
2sin(2B+C)cos(2B+C)=2cos(2B+C)cos(2B−C)
Assuming cos(2B+C)=0:
sin(2B+C)=cos(2B−C)
Since 2B+C=2π−2A, we have sin(2B+C)=cos(2A).
So, cos(2A)=cos(2B−C).
2. Simplify the target expression:
We want to find E=tan(2B)+cot(2B).
Using the identity tanθ+cotθ=cosθsinθ+sinθcosθ=sinθcosθsin2θ+cos2θ=sinθcosθ1=sin2θ2:
Given sinA=cosB+cosC, this specific structure in triangle problems often leads to an equilateral or specific triangle case. If we test an equilateral triangle where A=B=C=60∘:
sin60∘=23
cos60∘+cos60∘=21+21=1
Since 23=1, it is not equilateral.
However, using tan(2B)+cot(2B), the expression is simply 2cscB. For the identity to hold generally in such problems, the value simplifies to a constant. By testing the identity A=90∘,B+C=90∘, one can show the value is 2.
