If \( t \) is any real number, prove that [tex]1 + (\tan t)^2 = (\sec t)^2[/tex].
Answer (2)
1 + ( t an t ) 2 = ( sec t ) 2 L = 1 + ( cos t s in t ) 2 = 1 + co s 2 t s i n 2 t = co s 2 t co s 2 t + co s 2 t s i n 2 t = co s 2 t co s 2 t + s i n 2 = co s 2 t 1 = ( cos t 1 ) 2 = ( sec t ) 2 = R ==================================== t an x = cos x s in x s i n 2 x + co s 2 x = 1 sec x = cos x 1
We proved that 1 + tan 2 t = sec 2 t by using the definitions of tangent and secant and applying the Pythagorean identity. By rewriting the tangent in terms of sine and cosine, we demonstrated that the equation holds true. This relationship is fundamental in trigonometry and illustrates a key identity between these functions.
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