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| The absolutely free mathematics solutions | |||||||||||||||||||
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Welcome to the absolutely free mathematics solutions home page. We want to help you on your mathematics problems (elementary, secondary, senior high school, university students). We will response your problems as soon as possible and free of charge. Please do not hesitate to send your email to both [email protected] and cc: [email protected]. Your subject must be maths to make easy on processing. I wait your email. Thanks. | ||||||||||||||||||
| Example 1 | |||||||||||||||||||
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Proof that : tan 5A - tan 3A - tan 2A = tan 5A tan 3A tan 2A Proving : tan 5A = tan (3A + 2A) = (tan 3A + tan 2A)/(1 - tan 3A tan 2A) tan 5A (1 - tan 3A tan 2A) = tan 3A + tan 2A tan 5A - tan 5A tan 3A tan 2A = tan 3A + tan 2A tan 5A - tan 3A - tan 2A = tan 5A tan 3A tan 2A (qed) | ||||||||||||||||||
| Example 2 | |||||||||||||||||||
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Determine line equation through points P(3, 1) and Q(4, 5)! Solutions : (y - y1)/(y1 - y2) = (x - x1)/(x1 - x2) (y - 1)/(1 - 5) = (x - 3)/(3 - 4) (y - 1)/(-4) = (x - 3)/(-1) (y - 1)/4 = (x - 3) y - 1 = 4(x - 3) y = 1 + 4x - 12 y = 4x - 11 So the answer is : y = 4x - 11 | ||||||||||||||||||
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