Advertisements
Advertisements
प्रश्न
Without using trigonometric tables evaluate:
`(sin 65^@)/(cos 25^@) + (cos 32^@)/(sin 58^@) - sin 28^2. sec 62^@ + cosec^2 30^@`
Advertisements
उत्तर
`(sin 65^@)/(cos 25^@) + (cos 32^@)/(sin 58^@) - sin 28^2. sec 62^@ + cosec^2 30^@`
`= (sin(90^@ - 25^@))/cos 25^@ + (cos (90^@ - 58^@))/sin 58^@ - sin 28^@ xx 1/(cos(90^@ - 28^@) ) + 1/sin^2 30`
`= cos 25^@/cos 25^@ + (sin 58^@)/(sin 58^@) - sin 28^@ xx 1/sin 28^@ + (1/(1/2)^2)`
= 1 + 1 - 1 + 4
= 5
APPEARS IN
संबंधित प्रश्न
Without using trigonometric tables, evaluate the following:
`(\sin ^{2}20^\text{o}+\sin^{2}70^\text{o})/(\cos ^{2}20^\text{o}+\cos ^{2}70^\text{o}}+\frac{\sin (90^\text{o}-\theta )\sin \theta }{\tan \theta }+\frac{\cos (90^\text{o}-\theta )\cos \theta }{\cot \theta }`
Find the value of x, if sin x = sin 60° cos 30° – cos 60° sin 30°
Evaluate:
`(sin35^circ cos55^circ + cos35^circ sin55^circ)/(cosec^2 10^circ - tan^2 80^circ)`
Use tables to find cosine of 65° 41’
Prove that:
sec (70° – θ) = cosec (20° + θ)
If \[\frac{160}{3}\] \[\tan \theta = \frac{a}{b}, \text{ then } \frac{a \sin \theta + b \cos \theta}{a \sin \theta - b \cos \theta}\]
Sin 2A = 2 sin A is true when A =
A, B and C are interior angles of a triangle ABC. Show that
If ∠A = 90°, then find the value of tan`(("B+C")/2)`
Evaluate: 14 sin 30°+ 6 cos 60°- 5 tan 45°.
Sin 2B = 2 sin B is true when B is equal to ______.
