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

^{2}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 tan^{2} θ

= sin 30^{o} + √3 cos 30^{o} – 2 tan^{2} 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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