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If tan 35º = k, then the value oftan 145º – tan 1250/1+tan 145º tan 125 |
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Answer» Answer: 145° = 35°+35°+35°+35°+5° = 4 (35°) +5° = 4k+5 1250° = 35(35°) +25 = 35k+25 :.tan145° -tan1250° = 4k+5-(35k+25) = 4k+5-35k-25 =-31k -20 tan125°=3(35°) +20 = 3k+20 :.tan145° - tan1250°/1+tan145° tan125° = -31k-20 /(4k+5)(3k+20) |
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