Advertisements
Advertisements
Question
Express the following in terms of angles between 0° and 45°:
cosec68° + cot72°
Solution
cosec68° + cot72°
= cosec(90 - 22)° + cot(90 -18)°
= sec22° + tan18°
APPEARS IN
RELATED QUESTIONS
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 }`
Evaluate.
sin235° + sin255°
If A and B are complementary angles, prove that:
`(sinA + sinB)/(sinA - sinB) + (cosB - cosA)/(cosB + cosA) = 2/(2sin^2A - 1)`
Find A, if 0° ≤ A ≤ 90° and sin 3A – 1 = 0
Write the value of cos 1° cos 2° cos 3° ....... cos 179° cos 180°.
If θ is an acute angle such that \[\cos \theta = \frac{3}{5}, \text{ then } \frac{\sin \theta \tan \theta - 1}{2 \tan^2 \theta} =\] \[\cos \theta = \frac{3}{5}, \text{ then } \frac{\sin \theta \tan \theta - 1}{2 \tan^2 \theta} =\]
If x sin (90° − θ) cot (90° − θ) = cos (90° − θ), then x =
If θ is an acute angle such that sec2 θ = 3, then the value of \[\frac{\tan^2 \theta - {cosec}^2 \theta}{\tan^2 \theta + {cosec}^2 \theta}\]
If A, B and C are interior angles of a triangle ABC, then \[\sin \left( \frac{B + C}{2} \right) =\]
If y sin 45° cos 45° = tan2 45° – cos2 30°, then y = ______.
