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प्रश्न
Prove the following: sin58° sec32° + cos58° cosec32° = 2
बेरीज
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उत्तर
L.H.S.
= sin58° sec32° + cos58° cosec32°
= `sin(90° - 32°) xx (1)/(cos32°) + cos(90° - 32°) xx (1)/(sin32°)`
= `cos32° xx (1)/(cos32°) + sin32° xx (1)/(sin32°)`
= 1 + 1
= 2
= R.H.S.
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Trigonometric Equation Problem and Solution
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