Advertisements
Advertisements
प्रश्न
Evaluate the following: `(3sin37°)/(cos53°) - (5"cosec"39°)/(sec51°) + (4tan23° tan37° tan67° tan53°)/(cos17° cos67° "cosec"73° "cosec"23°)`
Advertisements
उत्तर
`(3sin37°)/(cos53°) - (5"cosec"39°)/(sec51°) + (4tan23° tan37° tan67° tan53°)/(cos17° cos67° "cosec"73° "cosec"23°)`
= `(3sin(90° - 53°))/(cos53°) - (5"cosec"(90° - 51°))/(sec51°) + (4tan(90° - 67°) tan(90° - 53°) xx 1/(cot67°) xx 1/(cot53°))/(cos(90° - 73°) cos(90° - 23°) xx 1/(sin73°) xx 1/(sin23°)`
= `(3cos53°)/(cos53°) - (5sec51°)/(sec51°) + (4 cos67° cos53° xx 1/(cot67°) xx 1/cot53°)/(sin73° sin23° xx 1/(sin73°) xx 1/sin23°)`
= 3 - 5 + 4
= 2.
APPEARS IN
संबंधित प्रश्न
Calculate the value of A, if (sin A - 1) (2 cos A - 1) = 0
If 2 sin x° − 1 = 0 and x° is an acute angle; find:
- sin x°
- x°
- cos x° and tan x°.
If 4 sin2 θ – 1 = 0 and angle θ is less than 90°, find the value of θ and hence the value of cos2 θ + tan2 θ.
Solve for x : cos (2x - 30°) = 0
If θ = 30°, verify that: sin2θ = `(2tanθ)/(1 ++ tan^2θ)`
If A = 30°, verify that cos2θ = `(1 - tan^2 θ)/(1 + tan^2 θ)` = cos4θ - sin4θ = 2cos2θ - 1 - 2sin2θ
In the given figure, PQ = 6 cm, RQ = x cm and RP = 10 cm, find
a. cosθ
b. sin2θ- cos2θ
c. Use tanθ to find the value of RQ
Evaluate the following: `(sec34°)/("cosec"56°)`
Evaluate the following: tan(78° + θ) + cosec(42° + θ) - cot(12° - θ) - sec(48° - θ)
If cos3θ = sin(θ - 34°), find the value of θ if 3θ is an acute angle.
