Question

if root 3 tan B = 3sinB,prove that sin^2-cos^2B =1/3

Trigonometry question!

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Answer 1

Step-by-step explanation:

√3 tan B = 3 SinB

Sin B / Cos B = √3 Sin B

Cos B = 1/√3

then

Sin B =✓1- Cos^2 B

Sin B = √2/√3

then now :

Sin^2 B - Cos^2 B = 2/3 - 1/3 = 1/3

Answer 2

Step-by-step explanation:

solution is-

√3tanB=3sinB

=> √3/3=sinB*cosB/sinB

=>1/√3„= cosB

=>cos^2B=1/3

Using sin square B +cos square B=1

sin^2B=1-cos^2B

sin^2B=1-(1/√3)^2

sin^2B= 1-1/3

sin^2B = 2/3

=> sin^2B-cos^2B=2/3-1/3

=>sin^2B-cos^2B=1/3

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