Come on! It's very simple

Geometry Level 4

sec ⁡ 1 1 ∘ sec ⁡ 1 9 ∘ − 2 cot ⁡ 7 1 ∘ \sec 11^\circ \sec 19^\circ - 2 \cot 71^\circ

2 tan ⁡ 1 9 ∘ \tan 19^\circ 2 sec ⁡ 1 9 ∘ \sec 19^\circ 2 cot ⁡ 1 1 ∘ \cot 11^\circ 2 sec ⁡ 1 1 ∘ \sec 11^\circ 2 tan ⁡ 1 1 ∘ \tan 11^\circ 2 cot ⁡ 1 9 ∘ \cot 19^\circ

Sachin Vishwakarma by Sachin Vishwakarma , 22, India

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1 solution

Simplify as: 2 sin ⁡ 3 0 ∘ cos ⁡ 1 1 ∘ cos ⁡ 1 9 ∘ − 2 tan ⁡ 1 9 ∘ \frac{2 \sin 30^\circ }{\cos 11^\circ \cos 19^\circ} -2\tan 19^\circ

Or 2 sin ⁡ ( 1 1 ∘ + 1 9 ∘ ) cos ⁡ 1 1 ∘ cos ⁡ 1 9 ∘ − 2 tan ⁡ 1 9 ∘ \frac {2 \sin( 11^\circ + 19 ^ \circ )}{\cos 11^\circ \cos 19^\circ} -2\tan 19^\circ

Or 2 tan ⁡ 1 1 ∘ + 2 tan ⁡ 1 9 ∘ − 2 tan ⁡ 1 9 ∘ 2\tan 11^\circ + 2 \tan 19^\circ -2 \tan 19^\circ

Or 2 tan ⁡ 1 1 ∘ 2\tan 11^\circ

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