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प्रश्न
Express the following in terms of angles between 0° and 45°:
cosec68° + cot72°
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उत्तर
cosec68° + cot72°
= cosec(90 - 22)° + cot(90 -18)°
= sec22° + tan18°
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'}`
Express sin 67° + cos 75° in terms of trigonometric ratios of angles between 0° and 45°
Evaluate.
`cot54^@/(tan36^@)+tan20^@/(cot70^@)-2`
Evaluate:
3cos80° cosec10° + 2 sin59° sec31°
Evaluate:
tan(55° - A) - cot(35° + A)
Prove that:
sec (70° – θ) = cosec (20° + θ)
If A and B are complementary angles, prove that:
cot A cot B – sin A cos B – cos A sin B = 0
Find A, if 0° ≤ A ≤ 90° and 4 sin2 A – 3 = 0
If A and B are complementary angles, then
If θ is an acute angle such that sec2 θ = 3, then the value of \[\frac{\tan^2 \theta - {cosec}^2 \theta}{\tan^2 \theta + {cosec}^2 \theta}\]
