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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 $\frac{\text{T}_3-\text{T}_5}{\text{T}_1}=\frac{\text{T}_5-\text{T}_7}{\text{T}_3}$
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Answer

We Have $\text{T}_\text{n}=\sin^\text{n}\text{x}+\cos^\text{n}\text{x}\cdots(\text{i})$ To show: $\frac{\text{T}_3-\text{T}_5}{\text{T}_1}=\frac{\text{T}_5-\text{T}_7}{\text{T}_3}$
$\text{L.H.S}=\frac{\text{T}_3-\text{T}_5}{\text{T}_1}$
$=\frac{(\sin^3\text{x}+\cos^3\text{x})-(\sin^5\text{x}+\cos^5\text{x})}{\sin\text{x}+\cos\text{x}}$ [Substituting the value of $T_3, T_5$ and $T_1$_ From (i)] $=\frac{\sin^3\text{x}-\sin^5\text{x}+\cos^3\text{x}-\cos^5\text{x}}{\sin\text{x}+\cos\text{x}}$
$=\frac{\sin^3\text{x}-(1-\sin^2\text{x})+\cos^3\text{x}(1-\cos^2\text{x})}{\sin\text{x}+\cos\text{x}}$
$=\frac{\sin^3\text{x}\cos^2\text{x}+\cos^3\text{x}\sin^2\text{x}}{\sin\text{x}+\cos\text{x}}$
$\Big[\because1-\sin^2\text{x}=\cos^2\text{x}\text{ and }1-\cos^2\text{x}=\sin^2\text{x}\Big]$
$=\frac{\sin^2\text{x}\cos^2\text{x}+(\sin\text{x}+\cos\text{x})}{\sin\text{x}+\cos\text{x}}$
$=\sin^2\text{x}\cos^2\text{x}$
$\text{R.H.S}=\frac{\sin^5\text{x}+\cos^5\text{x}-(\sin^7\text{x}+\cos^7\text{x})}{\sin^3\text{x}+\cos^3\text{x}}$
$=\frac{\sin^5\text{x}-\sin^7\text{x}+\cos^5\text{x}-\cos^7\text{x}}{\sin^3\text{x}+\cos^3\text{x}}$
$=\frac{\sin^5\text{x}(1-\sin^2\text{x})+\cos^5\text{x}(1-\cos^2\text{x})}{\sin^3\text{x}+\cos^3\text{x}}$
$=\frac{\sin^5\text{x}\cos^2\text{x}+\cos^5\text{x}\sin^2\text{x}}{\sin^3\text{x}+\cos^3\text{x}}$
$=\frac{\sin^5\text{x}\cos^2\text{x}(\sin^2\text{x}+\cos^3\text{x})}{\sin^2\text{x}+\cos^2\text{x}}$
$=\sin^2\text{x}\cos^2\text{x}$
$\text{L.H.S=R.H.S }$
$\text{Proved}$

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