Find the value of 2 tan2 θ + sin2 θ – 1.

If θ is an acute angle and sin θ = cos θ, find the value of 2 tan2 θ + sin2 θ – 1.

Answer :

Trigonometric Ratios Class 9 ICSE ML Aggarwal img 30

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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