Advertisements
Advertisements
प्रश्न
Evaluate the following
sec 50º sin 40° + cos 40º cosec 50º
उत्तर
We have,
sec50º sin40º + cos40º cosec50º
= sec(90º – 40º) sin40º + cos40º cosec(90º – 40º)
= cosec40º sin40º + cos40ºsec40º
`=\frac{\sin 40^\text{o}}{\sin 40^\text{o}}+\frac{\cos 40^\text{o}}{\cos40^\text{o}}`
= 1 + 1 = 2
APPEARS IN
संबंधित प्रश्न
Find the value of θ in each of the following :
(i) 2 sin 2θ = √3 (ii) 2 cos 3θ = 1
Evaluate the following :
`(sec 70^@)/(cosec 20^@) + (sin 59^@)/(cos 31^@)`
Express each one of the following in terms of trigonometric ratios of angles lying between 0° and 45°
cos 78° + sec 78°
Prove the following
sin θ sin (90° − θ) − cos θ cos (90° − θ) = 0
Evaluate: Cosec (65 + θ) – sec (25 – θ) – tan (55 – θ) + cot (35 + θ)
prove that:
sin (2 × 30°) = `(2 tan 30°)/(1+tan^2 30°)`
If A = B = 45° ,
show that:
sin (A - B) = sin A cos B - cos A sin B
prove that:
tan (2 x 30°) = `(2 tan 30°)/(1– tan^2 30°)`
Find the value of x in the following: 2 sin3x = `sqrt(3)`
`(2/3 sin 0^circ - 4/5 cos 0^circ)` is equal to ______.
