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If sec x = \(\sqrt{2}\) and \(\frac{3\pi}{2}\) < x <\(2\pi\) , find the values of \(\frac{1+tanx +cosec x}{1+cotx-cosecx}\) |
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Answer» Given, secx = \(\sqrt{2}\), \(\frac{3\pi}{2}<x<2\pi\) ∴ cos x = \(\frac{1}{\sqrt{2}}\), \(\frac{3\pi}{2}\) < x <\(2\pi\) and sinx = \(-\sqrt{1-cos^2x} \) as \(\frac{3\pi}{2}\) < x <\(2\pi\) = \(-\sqrt{1-(\frac{1}{\sqrt{2}})^2}\) = \(-\sqrt{\frac{1}{2}}\) =\(-\frac{1}{\sqrt{2}}\) ∴ cosecx = \(-\sqrt{2}\) Now, \(tanx=\frac{sinx}{cosx}=\frac{\frac{-1}{\sqrt{2}}}{\frac{1}{\sqrt{2}}}=-1\) ∴ \(cotx=\frac{1}{tanx}=-1\) Now, \(\frac{1+tanx+cosecx}{1+cotx-cosecx}=\frac{1+(-1)+(-\sqrt{2})}{1+(-1)-(-\sqrt{2})}=\frac{-\sqrt{2}}{2}=-1\) Which class you are from..?Is it 10 th level...Secx= root 2 So cox x = 1/ root2 So x= pi/4 or 45 degree Substitute the value of tan45 cot45 and cosec45 You will get root 2+1/root 2-1 Futher you msy simplfy by taking conjugate.... |
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