Advertisements
Advertisements
प्रश्न
Evaluate:
cosec (65° + A) – sec (25° – A)
Advertisements
उत्तर
cosec (65° + A) – sec (25° – A)
= cosec [90° – (25° – A)] – sec (25° – A)
= sec (25° – A) – sec (25° – A)
= 0
APPEARS IN
संबंधित प्रश्न
Express the trigonometric ratios sin A, sec A and tan A in terms of cot A.
Evaluate.
`(2tan53^@)/(cot37^@)-cot80^@/tan10^@`
Evaluate:
`(sin35^circ cos55^circ + cos35^circ sin55^circ)/(cosec^2 10^circ - tan^2 80^circ)`
If A and B are complementary angles, prove that:
cot B + cos B = sec A cos B (1 + sin B)
If \[\tan \theta = \frac{4}{5}\] find the value of \[\frac{\cos \theta - \sin \theta}{\cos \theta + \sin \theta}\]
If \[\tan A = \frac{3}{4} \text{ and } A + B = 90°\] then what is the value of cot B?
If \[\tan A = \frac{5}{12}\] \[\tan A = \frac{5}{12}\] find the value of (sin A + cos A) sec A.
If \[\tan \theta = \frac{3}{4}\] then cos2 θ − sin2 θ =
\[\frac{1 - \tan^2 45°}{1 + \tan^2 45°}\] is equal to
A triangle ABC is right-angled at B; find the value of `(sec "A". sin "C" - tan "A". tan "C")/sin "B"`.
