Sin ^{2} 60-tan 45+cos ^{2} 30-cot 90=ldots ldots ldots ldots

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8. Introduction to Trigonometry

easy

$\sin ^{2} 60-\tan 45+\cos ^{2} 30-\cot 90=\ldots \ldots \ldots \ldots$

A

$\frac{1}{2}$

B

$2$

C

$1$

D

$3$

Solution

$\sin ^{2} 60-\tan 45+\cos ^{2} 30-\cot 90=\left(\frac{\sqrt{3}}{2}\right)^{2}-1+\left(\frac{\sqrt{3}}{2}\right)^{2}-0$

$=\frac{3}{4}-1+\frac{3}{4}-0=\frac{3}{2}-1=\frac{1}{2}$

Standard 10

Mathematics

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