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Trigonometric Ratios Of Compounds — MATHS STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceMATHSTrigonometric Ratios Of Compounds4 Marks
Question
$\tan\text{x}\tan(\text{x}+\frac{\pi}{3})+\tan\text{x}(\frac{\pi}{3}-\text{x})\\+\tan(\text{x}+\frac{\pi}{3})\tan(\text{x}-\frac{\pi}{3})=-3$
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Answer

We have to prove that $\sin5\text{x}=5\cos^4\text{x}\sin\text{x}-10\cos^2\text{x}\sin^3\text{x}+\sin^5\text{x}$ $\text{LHS}=\sin\text{x}=\sin(3\text{x}+2\text{x})$ $=\sin3\text{x}\cos2\text{x}+\cos3\text{x}.\sin2\text{x}$ $=(3\sin\text{x}-4\sin^3\text{x})(2\cos^2\text{x}-1)\\+(4\cos^3\text{x}-3\cos\text{x})2\sin\text{x}\cos\text{x}.$ $=-3\sin\text{x}+4\sin^3\text{x}+6\sin\text{x}\cos^2\text{x}-8\sin^3\text{x}\cos^2\text{x}\\+8\cos^4\text{x}\sin\text{x}-6\cos^2\text{x}\sin\text{x}$ $=8\cos^4\text{x}\sin\text{x}-8\sin^3\text{x}\cos^2\text{x}-3\sin\text{x}+4\sin^3\text{x}$ $=5\cos^4\text{x}\sin\text{x}-10\sin^3\text{x}\cos^2\text{x}-3\sin\text{x}\\+3\cos^4\text{x}\sin\text{x}+4\sin^3\text{x}+2\sin^3\text{x}\cos^2\text{x}$ $=5\cos^4\text{x}\sin\text{x}-10\sin^3\text{x}\cos^2\text{x}\\-3\sin\text{x}(1-\cos^4​​\text{x})+2\sin^3\text{x}(2+\cos^2\text{x})$ $=5\cos^4\text{x}\sin\text{x}-10\sin^3\text{x}\cos^2\text{x}\\-3\sin\text{x}(1-\cos^2\text{x})(1+\cos^2\text{x})+2\sin^3\text{x}(2+\cos^2\text{x})$ $=5\cos^4\text{x}\sin\text{x}-10\sin^3\text{x}\cos^2\text{x}\\-3\sin^3\text{x}(1+\cos^2)+2\sin^3​​(2+\cos^2​​\text{x})$ $=5\cos^4\text{x}\sin\text{x}-10\sin^3\text{x}\cos^2\text{x}\\-\sin^3\text{x}\big[3(1+​​\cos^2​​​​\text{x})-2(2+\cos^2\text{x})\big]$ $=5\cos^4\text{x}\sin\text{x}-10\sin^3\text{x}\cos^2\text{x}\\-\sin^3\text{x}\big[3+3\cos^2​​\text{x}-4-2\cos^2\text{x}\big]$ $=5\cos^4\text{x}\sin\text{x}-10\sin^3\text{x}\cos^2\text{x}-\sin^3\text{x}\big[\cos^2​​\text{x}-1\big]$ $=5\cos^4\text{x}\sin\text{x}-10\sin^3\text{x}\cos^2\text{x}+\sin^5\text{x}$ $=\text{RHS}$

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