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1÷tanA - 1÷tan2A = cosec2A |
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Answer» -step explanation:1 − tan2A1 =csc2AL.H.S. = \dfrac{1}{\TAN A } -\dfrac{1}{\tan 2A } tanA1 − tan2A1 Using the trigonometric IDENTITY,\tan 2A = \dfrac{2\tan A}{1-\tan^2 A}tan2A= 1−tan 2 A2tanA = \dfrac{1}{\tan A } -\dfrac{1}{\dfrac{2\tan A}{1-\tan^2 A}} tanA1 − 1−tan 2 A2tanA 1 =\dfrac{1}{\tan A } -\dfrac{1-\tan^2 A}{2\tan A}= tanA1 − 2tanA1−tan 2 A TAKING LCM of denominator PART, we get=\dfrac{2-1+\tan^2 A}{2\tan A}= 2tanA2−1+tan 2 A =\dfrac{1+\tan^2 A}{2\tan A}= 2tanA1+tan 2 A =\dfrac{1}{\dfrac{2\tan A}{1+\tan^2 A} }= 1+tan 2 A2tanA 1 Using the trigonometric identity,\SIN 2A=\dfrac{2\tan A}{1+\tan^2 A}sin2A= 1+tan 2 A2tanA =\dfrac{1}{\sin 2A}= sin2A1 = \csc 2Acsc2A= R.H.S., proved.∴ \dfrac{1}{\tan A } -\dfrac{1}{\tan 2A } =\csc 2A tanA1 − tan2A1 =csc2A , proved |
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