Question

4 sin square 30 degree + cos square 60 degree minus 3 cos square 45 degree minus 10 square 45 degree is equal to 7 by 2​

Answer 1

Answer:

$4(sin^2{60}+cos^2{30})-3cos^2{45}-tan^2{45}=\frac{7}{2}$

Step-by-step explanation:

I think your question must be

Prove that:$4(sin^2{60}+cos^2{30})-3cos^2{45}-tan^2{45}=\frac{7}{2}$

Now,

$4(sin^2{60}+cos^2{30})-3cos^2{45}-tan^2{45}$

$=4[(\frac{\sqrt3}{2})^2+(\frac{\sqrt3}{2})^2]-3(\frac{1}{\sqrt2})^2-(1)^2$

$=4[\frac{3}{4}+\frac{3}{4}]-3(\frac{1}{2})-1$

$=4[\frac{6}{4}]-\frac{3}{2}-1$

$=6-\frac{5}{2}$

$=\frac{12-5}{2}$

$=\frac{7}{2}$

Answer 2

Hence proved $4(\sin^2{60}+\cos^2{30})-3\cos^2{45}-\tan^2{45}=\frac{7}{2}$

Step-by-step explanation:

To prove : $4(\sin^2{60}+\cos^2{30})-3\cos^2{45}-\tan^2{45}=\frac{7}{2}$

Proof :

Taking LHS,

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

Using trigonometric values,

$LHS=4[(\frac{\sqrt3}{2})^2+(\frac{\sqrt3}{2})^2]-3(\frac{1}{\sqrt2})^2-(1)^2$

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

$LHS=4[\frac{6}{4}]-\frac{3}{2}-1$

$LHS=6-\frac{5}{2}$

$LHS=\frac{12-5}{2}$

$LHS=\frac{7}{2}$

LHS=RHS

#Learn more

5 cos square 60 degree +4 sec square 30 degree - tan square 45 degree by sin square 30 degree +cos square 30 degree

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