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Evaluate:\(\frac{tan\, 35°}{cot\, 55°} + \frac{cot\, 78°}{tan\, 12°} - 1\) |
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Answer» We have, [∵ cot (90 – θ) = tan θ and tan (90 – θ) = cot θ] \(\frac{tan\, 35°}{cot\, 55°} + \frac{cot\, 78°}{tan\, 12°} - 1\) = \(\frac{tan\, (90°-\, 35°)}{cot\, 55°} + \frac{cot\,(90° -\, 12°)}{tan\, 12°} - 1\) = \(\frac{cot\, 55°}{cot\, 55°} + \frac{tan\, 12°}{tan\, 12°} - 1\) = 1 + 1 – 1 = 1 |
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