Advertisements
Advertisements
Question
If sin 3A = 1 and 0 < A < 90°, find `tan^2A - (1)/(cos^2 "A")`
Advertisements
Solution
sin 3A = 1
sin 3A = sin90°
3A = 90°
A = 30°
`tan^2A – (1)/(cos^2"A") = tan^2 30° – (1)/(cos^2 30°)`
= `(1/sqrt3)^2 – (1)/(sqrt3/2)^2`
= `(1)/(3) – (4)/(3)`
= `(–3)/(3)`
= – 1
APPEARS IN
RELATED QUESTIONS
Solve for x : cos (2x - 30°) = 0
Find the value of 'A', if cosec 3A = `(2)/sqrt(3)`
If θ = 15°, find the value of: cos3θ - sin6θ + 3sin(5θ + 15°) - 2 tan23θ
If θ < 90°, find the value of: `tan^2θ - (1)/cos^2θ`
Find lengths of diagonals AC and BD. Given AB = 24 cm and ∠BAD = 60°.
Evaluate the following: `(5sec68°)/("cosec"22°) + (3sin52° sec38°)/(cot51° cot39°)`
Express each of the following in terms of trigonometric ratios of angles between 0° and 45°: tan77° - cot63° + sin57°
If tan4θ = cot(θ + 20°), find the value of θ if 4θ is an acute angle.
If P, Q and R are the interior angles of ΔPQR, prove that `cot(("Q" + "R")/2) = tan "P"/(2)`
If secθ= cosec30° and θ is an acute angle, find the value of 4 sin2θ - 2 cos2θ.
