Find the value of sin θ + √3 cos θ – 2 tan2 θ.

If √2 tan 2θ = √6 and θ° < 2θ < 90°, find the value of sin θ + √3 cos θ – 2 tan² θ.

Answer :

√2 tan 2θ = √6

⇒ tan 2θ = √6/ √2

= √3

= tan 60^o

⇒ 2θ = 60^o

⇒ θ = 30^o

sin θ + √3 cos θ – 2 tan² θ

= sin 30^o + √3 cos 30^o – 2 tan² 30^o

= ½ + √3 x √3/2 – 2 (1/√3)²

= ½ + 3/2 – 2/3

= 4/2 – 2/3

= (12 – 4)/6

= 8/6

= 4/3

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