Advertisements
Advertisements
Question
If `sqrt(3)` sec 2θ = 2 and θ< 90°, find the value of
cos2 (30° + θ) + sin2 (45° - θ)
Advertisements
Solution
`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)`.
APPEARS IN
RELATED QUESTIONS
Solve for 'θ': `sin θ/(3)` = 1
If `sqrt(2) = 1.414 and sqrt(3) = 1.732`, find the value of the following correct to two decimal places tan60°
If θ < 90°, find the value of: `tan^2θ - (1)/cos^2θ`
In a right triangle ABC, right angled at C, if ∠B = 60° and AB = 15units, find the remaining angles and sides.
Find:
a. BC
b. AD
c. AC
Find the value of 'y' if `sqrt(3)` = 1.723.
Given your answer correct to 2 decimal places.
Evaluate the following: `(tan42°)/(cot48°) + (cos33°)/(sin57°)`
Evaluate the following: `(sin0° sin35° sin55° sin75°)/(cos22° cos64° cos58° cos90°)`
If cosθ = sin60° and θ is an acute angle find the value of 1- 2 sin2θ
Prove the following: `(tan(90° - θ)cotθ)/("cosec"^2 θ)` = cos2θ
