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प्रश्न
Evaluate the following: sin28° sec62° + tan49° tan41°
बेरीज
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उत्तर
sin28° sec62° + tan49° tan41°
= sin28° sec(90° - 28°) + tan49° tan(90° - 49°)
= sin28° cosec28° + tan49° cot49°
= `sin28° xx (1)/(sin28°) + tan49° xx (1)/(tan49°)`
= 1 + 1
= 2.
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Trigonometric Equation Problem and Solution
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