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प्रश्न
If `sqrt(3)` sec 2θ = 2 and θ< 90°, find the value of
cos2 (30° + θ) + sin2 (45° - θ)
उत्तर
`sqrt(3)`sec 2θ = 2
⇒ sec2θ = `(2)/sqrt(3)`
⇒ sec2θ = sec30°
⇒ 2θ = 30°
⇒ θ =15°
∴ cos2(30° + θ) + sin2(45° - θ)
= cos2(30° + 15°) + sin245° - 15°)
= cos245° sin230°
= `(1/sqrt(2))^2 + (1/2)^2`
= `(1)/(2) + (1)/(4)`
= `(3)/(4)`.
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