Question

If 15 tan²Ѳ + 4 sec²Ѳ = 23 find the value of taѲ + cotѲ

Answer 1

$15tan ^{2} \alpha + 4sec ^{2} \alpha = 23 \\ \\ 15tan ^{2} \alpha + 4 (1 + tan ^{2} \alpha ) = 23 \\ \\ 15tan ^{2} \alpha + 4 + 4 \: tan ^{2} \alpha = 23 \\ \\ 19tan ^{2} \alpha = 23 - 4 \\ \\ 19tan ^{2} \alpha = 19 \\ \\ tan ^{2} \alpha = \frac{19}{19} \\ \\ tan ^{2} \alpha = 1 \\ \\ tan \alpha = 1 \\ \\ so \\ \alpha = 45 \\ \\ now \: \\ \\ tan \alpha + cot \alpha \\ \\ = tan45 + cot45 \\ \\ = 1 + 1 \\ \\ = 2$

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