Sum of Trigonometric Ratios in Terms of Their Product
lf Sin2A+Co...
Question
lf Sin2A+Cos2B+Sin2C=1, then the triangle ABC is
A
isosceles
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B
equilateral
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C
right angled
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D
scalene
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Solution
The correct option is C right angled sin2A+cos2B=cos2C sin2A=(cosC−cosB)(cosC+cosB) sin2A=2×2cos(C+B2)cos(C−B)2sin(B+C2)sin(B−C2) sin2A=sin(B−C)sin(B+C) sinA=sin(B−C) A=B−C A+B+C=π 2B=π B=π2