Question

Prove this❤️❤️❤️
$\sqrt{ {sec}^{2} \: + {cosec}^{2} } \: = tan \: + cot$
​

Answer 1

Answer:

Hey mate i will help u

Refers with attachments

Answer 2

SOLUTION:-

L.H.S

$\sqrt{sec {}^{2} \theta + cosec {}^{2} \theta } \\ = > \sqrt{ \frac{1}{cos {}^{2} \theta } + \frac{1}{ {sin}^{2} \theta } } \\ \\ = > \sqrt{ \frac{ {sin}^{2} \theta + {cos}^{2} \theta}{ {sin}^{2} \theta \times cos {}^{2} \theta} } \\ \\ = > \sqrt{ \frac{1}{ {sin}^{2} \theta. {cos}^{2} \theta} } \\ \\ = > \frac{1}{sin \theta.cos \theta}$

R.H.S

$tan \theta \: + cot \theta \\ = > \frac{sin \theta}{cos \theta} + \frac{cos \theta}{sin \theta} \\ \\ = > \frac{ {sin}^{2} \theta + {cos}^{2} \theta}{sin \theta.cos \theta} \\ \\ = > \frac{1}{sin \theta.cos \theta}$

So, L.H.S = R.H.S

hope it helps ☺️