Advertisements
Advertisements
प्रश्न
A triangle ABC is right angles at B; find the value of`(secA.cosecC - tanA.cotC)/sinB`
Advertisements
उत्तर
Since, ABC is a right angled triangle, right angled at B.
So, A + C = 90°
`(secA.cosecC - tanA.cotC)/sinB`
= `(sec(90^circ - C).cosecC - tan(90^circ - C).cotC)/sin90^circ`
= `(cosecC.cosecC - cotC.cotC)/1`
= 1 ...[∵ cosec2θ – cot2θ = 1]
संबंधित प्रश्न
Evaluate.
sin(90° - A) cosA + cos(90° - A) sinA
Evaluate.
`(sin77^@/cos13^@)^2+(cos77^@/sin13^@)-2cos^2 45^@`
Show that : sin 42° sec 48° + cos 42° cosec 48° = 2
Evaluate:
`(sin35^circ cos55^circ + cos35^circ sin55^circ)/(cosec^2 10^circ - tan^2 80^circ)`
Use trigonometrical tables to find tangent of 17° 27'
Evaluate:
`(5sin66^@)/(cos24^@) - (2cot85^@)/(tan5^@)`
If 4 cos2 A – 3 = 0 and 0° ≤ A ≤ 90°, then prove that sin 3 A = 3 sin A – 4 sin3 A
The value of cosec(70° + θ) – sec(20° − θ) + tan(65° + θ) – cot(25° − θ) is
If cot( 90 – A ) = 1, then ∠A = ?
If sin 3A = cos 6A, then ∠A = ?
