Advertisements
Advertisements
प्रश्न
Evaluate cosec 31° − sec 59°
उत्तर
cosec 31° − sec 59° = cosec (90° − 59°) − sec 59°
= sec 59° − sec 59°
= 0
APPEARS IN
संबंधित प्रश्न
Without using trigonometric tables, evaluate the following:
`( i)\frac{\cos37^\text{o}}{\sin53^\text{o}}\text{ }(ii)\frac{\sin41^\text{o}}{\cos 49^\text{o}}(iii)\frac{\sin30^\text{o}17'}{\cos59^\text{o}\43'}`
If sec 4A = cosec (A− 20°), where 4A is an acute angle, find the value of A.
Prove the following trigonometric identities.
(cosecθ + sinθ) (cosecθ − sinθ) = cot2 θ + cos2θ
Solve.
`cos55/sin35+cot35/tan55`
Express the following in terms of angles between 0° and 45°:
cos74° + sec67°
Evaluate:
`(sin35^circ cos55^circ + cos35^circ sin55^circ)/(cosec^2 10^circ - tan^2 80^circ)`
Use tables to find sine of 10° 20' + 20° 45'
Use tables to find cosine of 8° 12’
Evaluate:
sin 27° sin 63° – cos 63° cos 27°
Evaluate:
`(cos75^@)/(sin15^@) + (sin12^@)/(cos78^@) - (cos18^@)/(sin72^@)`
\[\frac{1 - \tan^2 45°}{1 + \tan^2 45°}\] is equal to
If A, B and C are interior angles of a triangle ABC, then \[\sin \left( \frac{B + C}{2} \right) =\]
If \[\cos \theta = \frac{2}{3}\] then 2 sec2 θ + 2 tan2 θ − 7 is equal to
The value of \[\frac{\tan 55°}{\cot 35°}\] + cot 1° cot 2° cot 3° .... cot 90°, is
In the following figure the value of cos ϕ is
If ∆ABC is right angled at C, then the value of cos (A + B) is ______.
A triangle ABC is right-angled at B; find the value of `(sec "A". sin "C" - tan "A". tan "C")/sin "B"`.
Solve: 2cos2θ + sin θ - 2 = 0.
In ∆ABC, `sqrt(2)` AC = BC, sin A = 1, sin2A + sin2B + sin2C = 2, then ∠A = ? , ∠B = ?, ∠C = ?
