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| 1. |
Prove that following identitySin theeta- 2 sin theeta upon 2cos3 theeta - cos theeta = tan theeta |
| Answer» {tex}{\\sin \\theta - 2\\sin^3 \\theta \\over 2\\cos^3 \\theta -\\cos \\theta} = \\tan \\theta {/tex}Taking LHS,\xa0{tex}{\\sin \\theta - 2\\sin^3 \\theta \\over 2\\cos^3 \\theta -\\cos \\theta} \\\\= {\\sin \\theta (1-2\\sin^2 \\theta )\\over \\cos \\theta (2\\cos^2\\theta - 1)}\\\\= {\\sin \\theta (1-\\sin^2 \\theta -\\sin^2 \\theta )\\over \\cos \\theta (\\cos^2\\theta + \\cos^2\\theta - 1)}\\\\={\\sin \\theta (\\cos^2 \\theta -\\sin^2 \\theta )\\over \\cos \\theta (\\cos^2\\theta -(1- \\cos^2\\theta))}\\\\={\\sin \\theta (\\cos^2 \\theta -\\sin^2 \\theta )\\over \\cos \\theta (\\cos^2\\theta -\\sin^2\\theta)}\\\\{/tex}=\xa0{tex}\\tan \\theta = RHS {/tex}Hence Proved | |