If θ is an acute angle and sin θ = cos θ, find the value of 2 tan2 θ + sin2 θ – 1.
Answer :
Consider ∆ABC be right angled at B and ∠ACB = θ
sin θ = cos θ
⇒ sin θ/cos θ = 1
tan θ = AB/BC = 1
Take AB = x then BC = x
In right angled ∆ABC,
AC2 = AB2 + BC2
⇒ AC2 = x2 + x2 = 2x2
AC = √2x2
⇒ AC = (√2)x
In right angled ∆ABC
sin θ = perpendicular/hypotenuse
sin θ = AB/AC = x/√2x = 1/√2
Here
2 tan2 θ + sin2 θ – 1 = 2×(1)2 + (1/√2)2 – 1
= (2×1) + ½ – 1
= 2 + ½ – 1
= 1+ ½
Taking LCM
= (2 + 1)/2
= 3/2
Hence, 2 tan2 θ + sin2 θ – 1 = 3/2.
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