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Dulcie
Dulcie

Dulcie

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Dulcie
Asked: Tháng Mười 26, 20202020-10-26T08:40:41+00:00 2020-10-26T08:40:41+00:00In: Môn Toán

CMR a) cot ² a – cos ² a = cot ² a . cos ² a b) $\frac{1 + cos a }{sin a}$ = $\frac{sin a}{1 – cos a}$

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CMR
a) cot ² a – cos ² a = cot ² a . cos ² a
b) $\frac{1 + cos a }{sin a}$ = $\frac{sin a}{1 – cos a}$

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    1. Mít Mít

      Mít Mít

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      mit
      2020-10-26T08:41:41+00:00Added an answer on Tháng Mười 26, 2020 at 8:41 sáng

      a) $\cot^2a – \cos^2a$

      $= \dfrac{\cos^2a}{\sin^2a} – \cos^2a$

      $= \dfrac{\cos^2a – \sin^2a.\cos^2a}{\sin^2a}$

      $= \dfrac{\cos^2a(1 – \sin^2a)}{\sin^2a}$

      $= \dfrac{\cos^2a.\cos^2a}{\sin^2a}$

      $= \cot^2a.\cos^2a$

      b) $\dfrac{1 + \cos a}{\sin a} = \dfrac{\sin a}{1 – \cos a}$

      $\to (1 + \cos a)(1 – \cos a) = \sin^2a$

      $\to 1 – \cos^2a = \sin^2a$

      $\to \sin^2a = \sin^2a$ (hiển nhiên)

      Vậy $\dfrac{1 + \cos a}{\sin a} = \dfrac{\sin a}{1 – \cos a}$

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    2. Gerda

      Gerda

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      Gerda
      2020-10-26T08:42:05+00:00Added an answer on Tháng Mười 26, 2020 at 8:42 sáng

      a,

      $VT=\cot^2a-\cos^2a$

      $=\dfrac{\cos^2a}{\sin^2a}-\cos^2a$

      $=\dfrac{\cos^2a-\sin^2a\cos^2a}{\sin^2a}$

      $=\dfrac{\cos^2a(1-\sin^2a)}{\sin^2a}$

      $=\dfrac{\cos^2a.\cos^2a}{\sin^2a}$

      $=\cot^2a.\cos^2a$

      $=VP$

      b,

      $VT=\dfrac{1+\cos a}{\sin a}$

      $=\dfrac{(1+\cos a)(1-\cos a)}{\sin a(1-\cos a)}$

      $=\dfrac{1-\cos^2a}{\sin a(1-\cos a)}$

      $=\dfrac{\sin^2a}{\sin a(1-\cos a)}$

      $=\dfrac{\sin a}{1-\cos a}$

      $=VP$

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