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The value of \(\frac{sin\,19°}{cos\,71°}+\frac{cos\,73°}{sin\,17°}\) is equal to :(a) 0 (b) 1 (c) 2 (d) \(\frac12\) |
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Answer» (c) 2 Given exp. = \(\frac{sin(90°-71°)}{cos\,71°}+\frac{cox\,73°}{sin(90°-73°)}\) = \(\frac{cos\,71°}{cos\,71°}+\frac{cos\,73°}{cos\,73°}\) (∵sin (90° – θ) = cos θ) = 1 + 1 = 2. |
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