Question

Show that 2(cos^460+sin^430)-(tan ^260+cot^2 45)+3sec^2 30=1/4

Answer 1

hope this answer helps you

Answer 2

Step-by-step explanation:

To show : $2(\cos^4 60+\sin^4 30)-(\tan^2 60+\cot^2 45)+3\sec^2 30=\frac{1}{4}$

Solution :

Taking LHS,

$LHS=2(\cos^4 60+\sin^4 30)-(\tan^2 60+\cot^2 45)+3\sec^2 30$

Using trigonometric values,

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

$\sin 30=\frac{1}{2}$

$\tan 60=\sqrt3$

$\cot 45=1$

$\sec 30=\frac{2}{\sqrt3}$

Substitute all values,

$LHS=2((\frac{1}{2})^4+(\frac{1}{2})^4)-((\sqrt3)^2+(1)^2)+3(\frac{2}{\sqrt3})^2$

$LHS=2(\frac{1}{16}+\frac{1}{16})-(3+1)+3(\frac{4}{3})$

$LHS=2(\frac{2}{16})-(4)+4$

$LHS=\frac{1}{4}$

$LHS=RHS$

Hence proved.

#Learn more

Evaluate 2cos 2 60+3sec 2 30-2 tan 2 45 / sin 2 30 +cos 2 45

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