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Evaluate 2cos67°/sin23°-tan40°\cot50°-cos0°

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Evaluate 2cos67°/sin23°-tan40°\cot50°-cos0°

Posted by Ansu Bhadouriya 7 years, 8 months ago

CBSE > Class 10 > Mathematics

1 answers

Dinesh Kumar 7 years, 8 months ago

$\begin{array}{l}=\frac{2\cos67^o}{\sin23^o}-\frac{\tan{\displaystyle{\displaystyle40}^{{}^o}}}{cot50^o}-\cos0^o\\=\frac{2\sin(90-67)}{\sin{\displaystyle23}}-\frac{cot(90-40}{cot50}-1\\=\frac{2\sin23}{\sin{\displaystyle23}}-\frac{cot50}{cot50}-1\\=2-1-1\\=2-2\\=0\end{array}$

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CBSE > Class 10 > Mathematics