\frac{\sin150^\circ-5\cos300^\circ+7\tan225^\circ}{\tan135^\circ+3\sin210^\circ} =

AP ICET 2019 Shift 1 Paper

Question : 119 of 200

Marks: +1, -0

{\sin150°-5\cos300°+7\tan225°}/{\tan135°+3\sin210°}

Solution:

\;{\sin 150^{∘}-5 \cos 300^{∘}+7 \tan 225^{∘}}/{\tan 135^{∘}+3 \sin 210^{∘}}

=\;{\;{1}/{2}-{5}/{2}+7}/{-1-\;{3}/{2}}=\;{5}/{{-5}/{2}}=-2

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