Question

4 cot^2 45 - sec^2 60+ sin^2 60+cos^2 90

Answer 1

$4\cot^2 45-\sec^2 60+\sin^2 60+\cos^2 90$$=\dfrac{3}{4}$

Step-by-step explanation:

We have,

$4\cot^2 45-\sec^2 60+\sin^2 60+\cos^2 90$

To find, $4\cot^2 45-\sec^2 60+\sin^2 60+\cos^2 90=?$

∴ $4\cot^2 45-\sec^2 60+\sin^2 60+\cos^2 90$

$=4(1)^2-(2)^2+(\dfrac{\sqrt{3} }{2} )^2+(0)^2$

Since, $\cot 45=1, \sec 60=-2$ and

$sin 60=\dfrac{\sqrt{3} }{2}, \cos 90=0$

$=4\times 1-4+\dfrac{3 }{4}+0$

$=4-4+\dfrac{3 }{4}$

$=\dfrac{3}{4}$

Hence, $4\cot^2 45-\sec^2 60+\sin^2 60+\cos^2 90$$=\dfrac{3}{4}$