{"id":6823,"date":"2021-10-21T19:14:30","date_gmt":"2021-10-21T13:44:30","guid":{"rendered":"https:\/\/mathemerize.com\/?p=6823"},"modified":"2021-10-25T00:23:32","modified_gmt":"2021-10-24T18:53:32","slug":"solve-the-equation-2tan-12x1-cos-1x","status":"publish","type":"post","link":"https:\/\/mathemerize.com\/solve-the-equation-2tan-12x1-cos-1x\/","title":{"rendered":"Solve the equation : 2\\(tan^{-1}({2x+1})\\) = \\(cos^{-1}x\\)"},"content":{"rendered":"
Here, 2\\(tan^{-1}({2x+1})\\) = \\(cos^{-1}x\\)<\/p>\n
cos(2\\(tan^{-1}({2x+1})\\)) = x\u00a0 { We Know cos2x = \\({1-tan^2x\\over {1+tan^2x}}\\)}<\/p>\n
\\(\\therefore\\)\u00a0 \\({{1-{(2x+1)}^2}\\over {1-{(2x+1)}^2}}\\) = x\u00a0 \u00a0\\(\\implies\\)\u00a0 \u00a0(1 – 2x – 1)(1 + 2x + 1) = x(\\(4x^2 + 4x + 2\\))<\/p>\n
\\(\\implies\\)\u00a0 -2x.2(x + 1) = 2x(\\(2x^2 + 2x + 1\\))\u00a0 \\(\\implies\\)\u00a0 2x(\\(2x^2 + 2x + 1 + 2x + 2\\)) = 0<\/p>\n
\\(\\implies\\) x = 0\u00a0 or \\(2x^2 + 4x + 3\\) = 0 { No Solution }<\/p>\n
Verify\u00a0 x = 0<\/p>\n
\\(2tan^{-1}(1)\\) = \\(cos^{-1}(1)\\)\u00a0 \\(\\implies\\)\u00a0 \\(\\pi\\over 2\\) = \\(\\pi\\over 2\\)<\/p>\n
\\(\\therefore\\)\u00a0 x = 0 is only the solution.<\/p>\n
The value of \\(tan^{-1}(1)\\) + \\(cos^{-1}({-1\\over 2})\\) + \\(sin^{-1}({-1\\over 2})\\) is equal to<\/a><\/p>\n Prove that : \\(sin^{-1}{12\\over 13}\\) + \\(cot^{-1}{4\\over 3}\\) + \\(tan^{-1}{63\\over 16}\\) = \\(\\pi\\)<\/a><\/p>\n Evaluate \\(sin^{-1}(sin10)\\)<\/a><\/p>\n Prove that : \\(cos^{-1}{12\\over 13}\\) + \\(sin^{-1}{3\\over 5}\\) = \\(sin^{-1}{56\\over 65}\\)<\/a><\/p>\n