Question

please ans trignometry class 12th.Prove ​

Answer 1

Answer:

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Answer 2

Question:-Take LHS

The LHS can also be written that

$tan9 - tan27 - tan(90 - 27) + tan(90 - 9) \\ we \: know \: tan(90 - \alpha ) = cot \alpha \\ so \: using \: this \: we \: get \\ tan9 + cot9 - (tan27 + cot27) \\ convert \: into \: sin \: and \: cos \: term \: \\ \frac{sin9}{cos9} + \frac{cos9}{sin9} - ( \frac{sin27}{cos27} + \frac{cos27}{sin27}) \\ \frac{ {sin}^{2}9 + {cos}^{2}9 }{sin9cos9} - \frac{( {sin}^{2}27 + {cos}^{2}27) }{sin27cos27} \\ \frac{1}{sin9cos9} - \frac{1}{sin27cos27} \\ multiply \: numertor \: and \: denominator \:by \: 2 \\ \frac{2}{2sin9cos9} - \frac{2}{2sin27cos27} \\ we \: know \: 2sin \alpha cos \alpha = sin2 \alpha \\ \frac{2}{sin18} - \frac{2}{sin54} \\ sin18 = \frac{ \sqrt{5} - 1 }{4} and \: sin54 = \frac{ \sqrt{5} + 1}{4} \\ putting \: this \: value \: we \: get \\ \frac{2}{ \frac{ \sqrt{5} - 1}{4} } - \frac{2}{ \frac{ \sqrt{5} + 1 }{4} } \\ = \frac{8}{ \sqrt{5} - 1 } - \frac{8}{ \sqrt{5} + 1 } \\ = (8( \sqrt{5} + 1) - 8( \sqrt{5} - 1) \: ) \times \frac{1}{( \sqrt{5} + 1)( \sqrt{5} - 1) } \\ \frac{8 \sqrt{5} + 8 - 8 \sqrt{5} + 8 }{5 - 1} \\ \frac{16}{4} = 4 = LHS$

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