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Trigonometric Functions — MATHS STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceMATHSTrigonometric Functions4 Marks
Question
If $\text{T}_\text{n}=\sin^\text{n}\text{x}+\cos^\text{n}\text{x},$ Prove that $6\text{T}_{10}-15\text{T}_8+10\text{T}_6-1=0$
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Answer

$=-6\sin^2\text{x}\cos^2\text{x}(\sin^4\text{x}+\cos^4\text{x})+6\sin^4\cos^4\text{x}\\\ \ \ +9\sin^2\text{x}\cos^2\text{x}(\sin^4\text{x}+\cos^4\text{x})-3\sin^2\text{x}\cos^2\text{x}$ $=3\sin^2\text{x}\cos^2\text{x}(\sin^4\text{x}+\cos^4\text{x})=6\sin^4\text{x}\cos^4\text{x}-3\sin^2\text{x}\cos^2\text{x}$ $=3\sin^2\text{x}\cos^2\text{x}((\sin^2\text{x})^2+(\cos^2\text{x})^2\\\ \ +2\sin^2\text{x}\cos^2\text{x }2\sin^2\text{x}\cos^2)\\\ \ +6\sin^4\text{x}\cos^4\text{x}-3\sin^2\text{x}\cos^2\text{x}$ $($Adding and subtracting $2\sin^2\text{x}\cos^2\text{x})$$$ $=3\sin^2\text{x}\cos^2\text{x}((\sin^2\text{x}+\cos^2\text{x})^2-2\sin^2\text{x}\cos^2\text{x})\\\ \ +6\sin^4\text{x}\cos^4\text{x}-3\sin^2\text{x}\cos^2\text{x}$ $=3\sin^2\text{x}\cos^2\text{x}(1-2\sin^2\text{x}\cos^2\text{x})+6\sin^4\text{x}\cos^4\text{x}-3\sin^2\text{x}\cos^2\text{x}$ $=3\sin^2\text{x}\cos^2\text{x}-6\sin^4\text{x}\cos^4+6\sin^4\text{x}\cos^4\text{x}-3\sin^2\text{x}\cos^2\text{x}$ $=0$ $=\text{R.H.S}$ $\text{Proved}.$

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