(3cos260° + 2cot230° – 5sin245°) = ?


(3cos260° + 2cot230° – 5sin245°) = ?

(a) 1

(b) 4

(c) $\frac{17}{4}$

(d) $\frac{13}{6}$



As we know that,

$\sin 45^{\circ}=\frac{1}{\sqrt{2}}$

$\cos 60^{\circ}=\frac{1}{2}$

$\cot 30^{\circ}=\sqrt{3}$

By substituting these values, we get

$\left(3 \cos ^{2} 60^{\circ}+2 \cot ^{2} 30^{\circ}-5 \sin ^{2} 45^{\circ}\right)=3\left(\frac{1}{2}\right)^{2}+2(\sqrt{3})^{2}-5\left(\frac{1}{\sqrt{2}}\right)^{2}$





Hence, the correct option is (c).


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