Bài 1 Chứng minh rằng a)tan^2 a +1=1/cos^2a b)cot2a +1=1/sin^2a c)cos^4a-sin^4a=2cos^2a -1 giúp mình với mình cần gấp

0 bình luận về “Bài 1 Chứng minh rằng a)tan^2 a +1=1/cos^2a b)cot2a +1=1/sin^2a c)cos^4a-sin^4a=2cos^2a -1 giúp mình với mình cần gấp”

1. a,

   $VT= tan^2a+1$

   $= \dfrac{sin^2a}{cos^2a}+ 1$

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

   $= \dfrac{1}{cos^2a}= VP$

   b,

   $VT= cot^2a+1$

   $=\dfrac{cos^2a}{sin^2a}+1$

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

   $=\dfrac{1}{sin^2a}= VP$

   c,

   $VT= cos^4a-sin^4a$

   $= (cos^2a-sin^2a)(cos^2a+sin^2a)$

   $= cos^2a-sin^2a$

   $= cos^2a-(1-cos^2a)$

   $= 2cos^2a-1= VP$

   Bình luận

2. Đáp án:

   $\begin{array}{l}
   a){\tan ^2}a + 1\\
    = \dfrac{{{{\sin }^2}a}}{{{{\cos }^2}a}} + 1\\
    = \dfrac{{{{\sin }^2}a + {{\cos }^2}a}}{{{{\cos }^2}a}}\\
    = \dfrac{1}{{{{\cos }^2}a}}\\
   Vay\,{\tan ^2}a + 1 = \dfrac{1}{{{{\cos }^2}a}}\\
   b){\cot ^2}a + 1\\
    = \dfrac{{{{\cos }^2}a}}{{{{\sin }^2}a}} + 1\\
    = \dfrac{{{{\cos }^2}a + {{\sin }^2}a}}{{{{\sin }^2}a}}\\
    = \dfrac{1}{{{{\sin }^2}a}}\\
   c){\cos ^4}a – {\sin ^4}a\\
    = \left( {{{\cos }^2}a + {{\sin }^2}a} \right)\left( {{{\cos }^2}a – {{\sin }^2}a} \right)\\
    = 1.\left( {{{\cos }^2}a – \left( {1 – {{\cos }^2}a} \right)} \right)\\
    = 2{\cos ^2}a – 1
   \end{array}$

   Bình luận