Question

1+sin^2x=3sinxcosx then tan=?

Answer 1

Answer:

1 + sin²x = 3 sinx cosx 

(1 + sin²x)² = 9 sin²x cos²x 

1 + sin⁴x + 2 sin²x = 9 sin²x (1 - sin²x) 

1 + sin⁴x + 2 sin²x = 9sin²x - 9sin⁴x 

10sin⁴x - 7sin²x + 1 = 0 

sin²x = [ -(-7) ± √((-7)² - 4(10)(1))] / 2(10) 

sin²x = [ 7 ± √(49 - 40)] / 20 

sin²x = [ 7 ± 3] / 20 

sin²x = 1/2 or 1/5 

sinx = ±1/√2 or ±1/√5 

tanx 

= sinx/cosx 

= sinx/√(1 - sin²x) 

= (±1/√2)/√(1 - 1/2) OR (±1/√5)/√(1 - 1/5) 

= ±1 OR (±1/√5)/√(4/5) 

= ±1 OR ±1/2 

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Answer 2

Answer:

tan=1 or 1/2

Step-by-step explanation:

divide equation by cos^2

then we get equation as 2tan^2x-3tanx+1=0

by solving we get

tanx=1or 1/2