"DO NOT SEND SIMILAR QUERIES?" Answer (ii)
Q.4. Prove that
(ii). $4\left({\sin }^{4}30°+{\cos }^{4}60°\right)-3\left({\cos }^{2}45°-{\sin }^{2}90°\right)=2$

Dear Student,

Please find below the solution to the asked query:

We take L.H.S. and substitute trigonometric values :

$$\begin{gathered} 4\left({\mathrm{Sin}}^4 30° + {\mathrm{Cos}}^4 60°\right) - 3\left({\mathrm{Cos}}^2 45° - {\mathrm{Sin}}^2 90°\right) \\ \mathrm{We} \mathrm{know} : \mathrm{Sin} 30° = \mathrm{Cos} 60° =\frac{1}{2} , \mathrm{Cos} 45° = \frac{1}{\sqrt{2}} \mathrm{and} \mathrm{Sin} 90° =1 , \mathrm{So} \\ \Rightarrow 4\left({\left(\frac{1}{2}\right)}^4 + {\left(\frac{1}{2}\right)}^4\right) - 3\left({\left(\frac{1}{\sqrt{2}}\right)}^2 - 1^2\right) \\ \Rightarrow 4\left(\frac{1}{16}+\frac{1}{16}\right) - 3\left(\frac{1}{2} -1\right) \\ \Rightarrow 4\left(\frac{2}{16}\right) - 3\left(\frac{1 - 2}{2}\right) \\ \Rightarrow 4\left(\frac{1}{8}\right) - 3\left(-\frac{1}{2}\right) \\ \Rightarrow \frac{1}{2}+\frac{3}{2} \\ \Rightarrow \frac{4}{2} \\ \mathbf{\Rightarrow }\mathbf{2} \end{gathered}$$
Therefore,

L.H.S. = R.H.S.                                       ( Hence proved )


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