Advertisements
Advertisements
प्रश्न
The value of cos 1° cos 2° cos 3° ..... cos 180° is
विकल्प
1
0
−1
None of these
उत्तर
Here we have to find: `cos 1° cos 2° cos 3°........... cos180°`
`cos 1° cos 2° cos 3°................ cos180°`
`= cos 1° cos 2° cos 3°............. cos 89° cos 90° cos 91° ............. os 180°` ` [since cos 90°=0]`
`= cos 1° cos 2° cos 3°............0xx cos 90° cos180°`
`= cos 1° cos 2° cos 3°........0 xxcos 90°........... cos 180°`
`= 0`
APPEARS IN
संबंधित प्रश्न
If A, B, C are the interior angles of a triangle ABC, prove that `\tan \frac{B+C}{2}=\cot \frac{A}{2}`
Evaluate `(tan 26^@)/(cot 64^@)`
Prove the following trigonometric identities.
(cosecθ + sinθ) (cosecθ − sinθ) = cot2 θ + cos2θ
if `cosec A = sqrt2` find the value of `(2 sin^2 A + 3 cot^2 A)/(4(tan^2 A - cos^2 A))`
Evaluate.
`cot54^@/(tan36^@)+tan20^@/(cot70^@)-2`
Use tables to find the acute angle θ, if the value of sin θ is 0.4848
Use tables to find the acute angle θ, if the value of sin θ is 0.3827
Use tables to find the acute angle θ, if the value of tan θ is 0.4741
Evaluate:
sin 27° sin 63° – cos 63° cos 27°
Prove that:
sec (70° – θ) = cosec (20° + θ)
If 4 cos2 A – 3 = 0 and 0° ≤ A ≤ 90°, then prove that sin 3 A = 3 sin A – 4 sin3 A
If 0° < A < 90°; find A, if `(cos A )/(1 - sin A) + (cos A)/(1 + sin A) = 4`
If \[\cos \theta = \frac{2}{3}\] find the value of \[\frac{\sec \theta - 1}{\sec \theta + 1}\]
Write the acute angle θ satisfying \[\cos B = \frac{3}{5}\]
If x tan 45° cos 60° = sin 60° cot 60°, then x is equal to
If θ is an acute angle such that sec2 θ = 3, then the value of \[\frac{\tan^2 \theta - {cosec}^2 \theta}{\tan^2 \theta + {cosec}^2 \theta}\]
The value of
If 5θ and 4θ are acute angles satisfying sin 5θ = cos 4θ, then 2 sin 3θ −\[\sqrt{3} \tan 3\theta\] is equal to
The value of tan 1° tan 2° tan 3°…. tan 89° is
Prove that `"tan A"/"cot A" = (sec^2"A")/("cosec"^2"A")`
