Advertisements
Advertisements
प्रश्न
If θ < 90°, find the value of: sin2θ + cos2θ
उत्तर
Since θ <90°,
Consider θ = 45°
∴ sin2 + cos2
= sin245° + cos245°
= `(1/sqrt(2))^2 + (1/sqrt(2))^2`
= `(1)/(2) + (1)/(2)`
= 1.
APPEARS IN
संबंधित प्रश्न
If sin 3A = 1 and 0 < A < 90°, find sin A
In ΔABC, ∠B = 90° , AB = y units, BC = `(sqrt3)` units, AC = 2 units and angle A = x°, find:
- sin x°
- x°
- tan x°
- use cos x° to find the value of y.
If A = B = 60°, verify that: sin(A - B) = sinA cosB - cosA sinB
If θ < 90°, find the value of: `tan^2θ - (1)/cos^2θ`
In the given figure, AB and EC are parallel to each other. Sides AD and BC are 1.5 cm each and are perpendicular to AB. Given that ∠AED = 45° and ∠ACD = 30°. Find:
a. AB
b. AC
c. AE
Find the value 'x', if:
Find x and y, in each of the following figure:
Evaluate the following: `(sin25° cos43°)/(sin47° cos 65°)`
If cosθ = sin60° and θ is an acute angle find the value of 1- 2 sin2θ
If secθ= cosec30° and θ is an acute angle, find the value of 4 sin2θ - 2 cos2θ.
