Advertisements
Advertisements
Question
Find the sine ratio of θ in standard position whose terminal arm passes through (3, 4)
Solution
Terminal arm passes through (3, 4).
Hence,
APPEARS IN
RELATED QUESTIONS
`(\text{i})\text{ }\frac{\cot 54^\text{o}}{\tan36^\text{o}}+\frac{\tan 20^\text{o}}{\cot 70^\text{o}}-2`
Express each of the following in terms of trigonometric ratios of angles between 0º and 45º;
(i) cosec 69º + cot 69º
(ii) sin 81º + tan 81º
(iii) sin 72º + cot 72º
Evaluate `(tan 26^@)/(cot 64^@)`
Evaluate cosec 31° − sec 59°
Prove the following trigonometric identities.
(cosecA − sinA) (secA − cosA) (tanA + cotA) = 1
Evaluate.
`cot54^@/(tan36^@)+tan20^@/(cot70^@)-2`
Evaluate.
`cos^2 26^@+cos65^@sin26^@+tan36^@/cot54^@`
Evaluate:
`2 tan57^circ/(cot33^circ) - cot70^circ/(tan20^circ) - sqrt(2) cos45^circ`
Find the value of x, if sin 2x = 2 sin 45° cos 45°
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'
If A and B are complementary angles, then
tan 5° ✕ tan 30° ✕ 4 tan 85° is equal to
Prove the following.
tan4θ + tan2θ = sec4θ - sec2θ
Evaluate: `3(sin72°)/(cos18°) - (sec32°)/("cosec"58°)`.
Evaluate: `2(tan57°)/(cot33°) - (cot70°)/(tan20°) - sqrt(2) cos 45°`
In the case, given below, find the value of angle A, where 0° ≤ A ≤ 90°.
cos(90° - A) · sec 77° = 1
The value of tan 72° tan 18° is
If sin 3A = cos 6A, then ∠A = ?
2(sin6 θ + cos6 θ) – 3(sin4 θ + cos4 θ) is equal to ______.
