Prove the following: sin58° sec32° + cos58° cosec32° = 2

Sum

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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Chapter 22 Trigonometrical Ratios of Standard Angles
Exercise 27.3 | Q 14.2

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