Advertisements
Advertisements
Question
Express the following in term of angles between 0° and 45° :
cos 74° + sec 67°
Advertisements
Solution
cos 74° + sec 67°
= cos(90° – 16°) + sec(90° – 23°)
= sin 16° + cosec 23°
APPEARS IN
RELATED QUESTIONS
Without using trigonometric tables evaluate:
`(sin 65^@)/(cos 25^@) + (cos 32^@)/(sin 58^@) - sin 28^2. sec 62^@ + cosec^2 30^@`
if `cot theta = 1/sqrt3` find the value of `(1 - cos^2 theta)/(2 - sin^2 theta)`
Show that : sin 42° sec 48° + cos 42° cosec 48° = 2
Evaluate:
`sin80^circ/(cos10^circ) + sin59^circ sec31^circ`
Evaluate:
14 sin 30° + 6 cos 60° – 5 tan 45°
A triangle ABC is right angles at B; find the value of`(secA.cosecC - tanA.cotC)/sinB`
Use tables to find cosine of 8° 12’
Use tables to find cosine of 26° 32’
Use tables to find cosine of 9° 23’ + 15° 54’
Evaluate:
`(cos75^@)/(sin15^@) + (sin12^@)/(cos78^@) - (cos18^@)/(sin72^@)`
If A and B are complementary angles, prove that:
cosec2 A + cosec2 B = cosec2 A cosec2 B
Find A, if 0° ≤ A ≤ 90° and 4 sin2 A – 3 = 0
If 0° < A < 90°; find A, if `sinA/(secA - 1) + sinA/(secA + 1) = 2`
If A + B = 90° and \[\cos B = \frac{3}{5}\] what is the value of sin A?
Prove that :
tan5° tan25° tan30° tan65° tan85° = \[\frac{1}{\sqrt{3}}\]
Evaluate: `(cot^2 41°)/(tan^2 49°) - 2 (sin^2 75°)/(cos^2 15°)`
A triangle ABC is right-angled at B; find the value of `(sec "A". sin "C" - tan "A". tan "C")/sin "B"`.
If sin 3A = cos 6A, then ∠A = ?
In the given figure, if AB = 14 cm, BD = 10 cm and DC = 8 cm, then the value of tan B is ______.
The value of the expression (cos2 23° – sin2 67°) is positive.
