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प्रश्न
Express the following in terms of angle between 0° and 45°:
sin 59° + tan 63°
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उत्तर
sin 59° + tan 63°
= sin(90 - 31)° + tan(90 - 27)°
= cos 31° + cot 27°
संबंधित प्रश्न
Express the following in terms of angles between 0° and 45°:
cosec68° + cot72°
Evaluate:
`3 sin72^circ/(cos18^circ) - sec32^circ/(cosec58^circ)`
Evaluate:
`(cot^2 41^circ)/(tan^2 49^circ) - 2 sin^2 75^circ/cos^2 15^circ`
Find the value of angle A, where 0° ≤ A ≤ 90°.
sin (90° – 3A) . cosec 42° = 1
Evaluate:
cos 40° cosec 50° + sin 50° sec 40°
Prove that:
tan (55° - A) - cot (35° + A)
Prove that:
sec (70° – θ) = cosec (20° + θ)
If θ is an acute angle such that \[\cos \theta = \frac{3}{5}, \text{ then } \frac{\sin \theta \tan \theta - 1}{2 \tan^2 \theta} =\] \[\cos \theta = \frac{3}{5}, \text{ then } \frac{\sin \theta \tan \theta - 1}{2 \tan^2 \theta} =\]
Express the following in term of angles between 0° and 45° :
cos 74° + sec 67°
Evaluate:
3 cos 80° cosec 10°+ 2 sin 59° sec 31°
