Question

Sin square 45 degree +cos square 45 degree by tan square 60 degree

Answer 1

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Answer 2

$\dfrac{\sin^2 45^\circ+\cos^2 45^\circ}{\tan^2 60^\circ}=\dfrac{1}{3}$

Step-by-step explanation:

Given : Expression $\dfrac{\sin^2 45^\circ+\cos^2 45^\circ}{\tan^2 60^\circ}$

To find : Solve the expression ?

Solution :

Expression $\dfrac{\sin^2 45^\circ+\cos^2 45^\circ}{\tan^2 60^\circ}$

We know by trigonometric values,

$\sin 45=\frac{1}{\sqrt2}$, $\cos 45=\frac{1}{\sqrt2}$ and $\tan 60=\sqrt3$

Substitute the values in the expression,

$\dfrac{\sin^2 45^\circ+\cos^2 45^\circ}{\tan^2 60^\circ}=\dfrac{(\sin45)^2+(\cos 45)^2}{(\tan 60)^2}$

$\dfrac{\sin^2 45^\circ+\cos^2 45^\circ}{\tan^2 60^\circ}=\dfrac{(\frac{1}{\sqrt2})^2+(\frac{1}{\sqrt2})^2}{(\sqrt3)^2}$

$\dfrac{\sin^2 45^\circ+\cos^2 45^\circ}{\tan^2 60^\circ}=\dfrac{\frac{1}{2}+\frac{1}{2}}{3}$

$\dfrac{\sin^2 45^\circ+\cos^2 45^\circ}{\tan^2 60^\circ}=\dfrac{\frac{2}{2}}{3}$

$\dfrac{\sin^2 45^\circ+\cos^2 45^\circ}{\tan^2 60^\circ}=\dfrac{1}{3}$

#Learn more

Sin square 30 degree sin square 45 degree sin square 60 degree sin square 90 degree

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