Question

2tan20°\ cot70° + 3cot65°\2tan25°​

Answer 1

Solution :

$\implies\tt \frac{2 \tan20 \degree }{ \cot70\degree} + \frac{3 \cot65\degree }{2 \tan25\degree } \\ \\\implies \tt \frac{2 \tan20 \degree }{ \cot(90 \degree - 20\degree)} + \frac{3 \cot(90 \degree - 25)\degree }{2 \tan25\degree } \\ \\\implies \tt \frac{2 \tan20 \degree }{ \tan20\degree} + \frac{3 \tan25\degree }{2 \tan25\degree } \\ \\\implies \tt \tan20\degree + \tan25\degree \\ \\\implies \tt \tan(20 + 25) \\ \\ \large \boxed{\tt \green{ \tan45 \degree = 1}}$

Important identities :

$\implies \tt \sin(90 - \theta) = \cos \theta \\ \\\implies \tt \cos(90 - \theta) = \sin \theta \\ \\\implies \tt \csc(90 - \theta) = \sec \theta \\ \\\implies \tt \sec(90 - \theta) = \csc \theta \\ \\\implies \tt \tan(90 - \theta) = \cot \theta \\ \\\implies \tt \cot(90 - \theta) = \tan \theta \\ \\ \tt \implies {\sin}^{2} +{\cos}^{2}=1 \\ \\\tt \implies {\tan}^{2} + 1 = {\sec}^{2} \\ \\ \tt \implies 1 + {\cot}^{2} = {\csc}^{2}$