Question

tan cube alpha upon one plus tan square alpha +cot cube alpha upon oneplus cot alpha =Sec alpha into cos alpha - 2 sin alpha cos alpha​

Answer 1

$\huge\color{Black}\boxed{\colorbox{yellow}{❀Answer❀}}$

$\frac{ {tan}^{3} \alpha }{1 + {tan}^{2} \alpha } + \frac{ {cot}^{3} \alpha }{1 + {cot}^{2} \alpha } \\ = \frac{ {tan}^{3} \alpha }{ {sec}^{2} \alpha } + \frac{ {cot}^{3} \alpha }{ {cosec}^{2} \alpha } \\ = \frac{ {sin}^{3} \alpha }{ {cos}^{3} \alpha } \times {cos}^{2} \alpha + \frac{ {cos}^{3} \alpha }{ {sin}^{3} \alpha } \times {sin}^{2} \alpha \\ = \frac{ {sin}^{3} \alpha }{cos \alpha } + \frac{ {cos}^{3} \alpha }{sin \alpha } \\ = \frac{ {sin}^{4} \alpha + {cos}^{4} \alpha }{sin \alpha cos \alpha } \\ = \frac{( { {sin}^{2} \alpha + {cos}^{2} \alpha })^{2} - 2 {sin}^{2} \alpha {cos}^{2} \alpha }{sin \alpha cos \alpha} \\ = \frac{1 - 2 {sin}^{2} \alpha {cos}^{2} \alpha }{sin \alpha cos \alpha } \\ = \frac{1}{sin \alpha \: cos \alpha } - \frac{2 {sin}^{2} \alpha {cos}^{2} \alpha }{sin \alpha cos \alpha } \\ = sec \alpha cosec \alpha - 2sin \alpha cos \alpha$