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3.Trigonometrical Ratios, Functions and Identities
easy
$\frac{{\cot 54^\circ }}{{\tan 36^\circ }} + \frac{{\tan 20^\circ }}{{\cot 70^\circ }}$ =
A
$2$
B
$3$
C
$1$
D
$0$
Solution
(a) $\tan (90^\circ – \theta ) = \cot \theta ,\;\cot (90^\circ – \theta ) = \tan \theta .$
Therefore $\frac{{\cot 54^\circ }}{{\tan 36^\circ }} + \frac{{\tan 20^\circ }}{{\cot 70^\circ }}$
$ = \frac{{\cot 54^\circ }}{{\tan (90^\circ – 54^\circ )}} + \frac{{\tan 20^\circ }}{{(\cot 90^\circ – 20^\circ )}}$
$\frac{{\cot 54^\circ }}{{\cot 54^\circ }} + \frac{{\tan 20^\circ }}{{\tan 20^\circ }} = 1 + 1 = 2$.
Standard 11
Mathematics
