Advertisements
Advertisements
प्रश्न
Show that : sin 42° sec 48° + cos 42° cosec 48° = 2
Advertisements
उत्तर
sin 42° sec 48° + cos 42° cosec 48° = 2
Consider sin 42° sec 48° + cos 42° cosec 48°
`=>` sin 42° sec (90° – 42°) + cos 42° cosec (90° – 42°)
`=>` sin 42° cosec 42° + cos 42° sec 42°
`=> sin 42^circ xx 1/(sin42^circ) + cos42^circ xx 1/(cos42^circ)`
`=>` 1 + 1 = 2
संबंधित प्रश्न
Evaluate cosec 31° − sec 59°
Write all the other trigonometric ratios of ∠A in terms of sec A.
Prove the following trigonometric identities.
(secθ + cosθ) (secθ − cosθ) = tan2θ + sin2θ
Evaluate:
`sin80^circ/(cos10^circ) + sin59^circ sec31^circ`
Evaluate:
cosec (65° + A) – sec (25° – A)
Find the value of x, if sin 2x = 2 sin 45° cos 45°
If 5 tan θ − 4 = 0, then the value of \[\frac{5 \sin \theta - 4 \cos \theta}{5 \sin \theta + 4 \cos \theta}\] is:
Prove that :
tan5° tan25° tan30° tan65° tan85° = \[\frac{1}{\sqrt{3}}\]
In ∆ABC, `sqrt(2)` AC = BC, sin A = 1, sin2A + sin2B + sin2C = 2, then ∠A = ? , ∠B = ?, ∠C = ?
Prove the following:
tan θ + tan (90° – θ) = sec θ sec (90° – θ)
