Advertisements
Advertisements
प्रश्न
Evaluate:
`(3sin72^@)/(cos18^@) - sec32^@/(cosec58^@)`
उत्तर
`(3sin72^@)/(cos18^@) - sec32^@/(cosec58^@)`
= `(3sin(90^@ - 18^@))/(cos18^@) - (sec(90^@ - 58^@))/(cosec58^@)`
= `(3cos18^@)/(cos18^@) - (cosec58^@)/(cosec58^@)`
= 3 – 1
= 2
APPEARS IN
संबंधित प्रश्न
Prove the following trigonometric identities.
(secθ + cosθ) (secθ − cosθ) = tan2θ + sin2θ
Evaluate.
sin(90° - A) cosA + cos(90° - A) sinA
Show that : sin 42° sec 48° + cos 42° cosec 48° = 2
Find the value of x, if tan x = `(tan60^circ - tan30^circ)/(1 + tan60^circ tan30^circ)`
Use trigonometrical tables to find tangent of 42° 18'
Write the acute angle θ satisfying \[\cos B = \frac{3}{5}\]
Write the value of cos 1° cos 2° cos 3° ....... cos 179° cos 180°.
Write the value of tan 10° tan 15° tan 75° tan 80°?
Evaluate: `(sin 80°)/(cos 10°)`+ sin 59° sec 31°
In the given figure, if AB = 14 cm, BD = 10 cm and DC = 8 cm, then the value of tan B is ______.
