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question ccs2 330 —ccs 57 690 Sin

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Please find below the solution to the asked query:

$$\begin{gathered} \mathrm{L}.\mathrm{H}.\mathrm{S}.=\frac{{\cos }^{2}33-{\cos }^{2}57}{{\sin }^2{\displaystyle \frac{21}{2}}-{\sin }^2{\displaystyle \frac{69}{2}}} \\ =\frac{{\sin }^{2}57-{\sin }^{2}33}{{\sin }^2{\displaystyle \frac{21}{2}}-{\sin }^2{\displaystyle \frac{69}{2}}} \left[\mathrm{As} \cos \left(\mathrm{A}\right)=\sin \left(90-\mathrm{A}\right)\right] \\ \mathrm{Use} {\sin }^2\mathrm{A}-{\sin }^2\mathrm{B}=\sin \left(\mathrm{A}+\mathrm{B}\right)\sin \left(\mathrm{A}-\mathrm{B}\right) \\ \mathrm{L}.\mathrm{H}.\mathrm{S}.=\frac{\sin \left(57+33\right)\sin \left(57-33\right)}{\sin \left({\displaystyle \frac{21}{2}}+{\displaystyle \frac{69}{2}}\right)\sin \left(\frac{21}{2}-\frac{69}{2}\right)} \\ =\frac{\sin \left(90\right)\sin \left(24\right)}{\sin \left({\displaystyle \frac{90}{2}}\right)\sin \left(-\frac{48}{2}\right)} \\ =\frac{\sin \left(90\right)\sin \left(24\right)}{\sin \left({\displaystyle 45}\right)\sin \left(-24\right)} \\ =-\frac{\sin \left(90\right)\sin \left(24\right)}{\sin \left({\displaystyle 45}\right)\sin \left(24\right)} \left[\mathrm{As} \sin \left(-\mathrm{A}\right)=-\sin \left(\mathrm{A}\right)\right] \\ =-\frac{\sin \left(90\right)}{\sin \left(45\right)} \\ =-\frac{1}{{\displaystyle \frac{1}{\sqrt{2}}}} \\ =-\sqrt{2} =\mathrm{R}.\mathrm{H}.\mathrm{S}. \end{gathered}$$

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