3tan^2theta-1=0;findcos^2theta
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3 tan²ø - 1 = 0
3 tan²ø = 1
tan²ø = 1/3
tanø = ±(1/√3) = ±(tan 30°) or ±(tan π/6)
ø = ±30°
cos²ø = [cos (±30°)]² = (cos 30°)²
= (√3/2)² = 3/4
So, cos²ø = 3/4
Thankyou!!!
3 tan²ø = 1
tan²ø = 1/3
tanø = ±(1/√3) = ±(tan 30°) or ±(tan π/6)
ø = ±30°
cos²ø = [cos (±30°)]² = (cos 30°)²
= (√3/2)² = 3/4
So, cos²ø = 3/4
Thankyou!!!
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