Proof-that-4-cos-4-a-sin-4-a-cos-4-a-sin-4-a-3-cos-4a-sec-2a-

Question Number 116891 by bemath last updated on 07/Oct/20

Proof that \frac{4 (\cos^{4} (a) + \sin^{4} (a))}{\cos^{4} (a) - \sin^{4} (a)} = (3 + \cos (4 a)) \sec (2 a)

Answered by john santu last updated on 07/Oct/20

\Rightarrow \frac{4 {{(\sin^{2} a + \cos^{2} a)}^{2} - 2 \sin^{2} a \cos^{2} a}}{\cos^{2} a - \sin^{2} a} =

\frac{4 (1 - \frac{1}{2} \sin^{2} 2 a)}{\cos 2 a} = \frac{4 - 2 \sin^{2} 2 a}{\cos 2 a}

= \frac{4 - 2 (\frac{1}{2} - \frac{1}{2} \cos 4 a)}{\cos 2 a} = (3 + \cos 4 a) \sec 2 a

p r o v e d

Answered by Dwaipayan Shikari last updated on 07/Oct/20

\frac{4 ({c o s}^{4} a + {s i n}^{4} a)}{({c o s}^{2} a + {s i n}^{2} a) ({c o s}^{2} a - {s i n}^{2} a)}

= \frac{4 ({({c o s}^{2} a + {s i n}^{2} a)}^{2} - 2 {s i n}^{2} {a c o s}^{2} a)}{c o s 2 a}

= \frac{4 (1 - \frac{1}{2} {s i n}^{2} 2 a)}{c o s 2 a} = \frac{4 - 2 {s i n}^{2} 2 a}{c o s 2 a} = \frac{4 - 1 + c o s 4 a}{c o s 2 a} = (3 + c o s 4 a) s e c 2 a