$\tan(A + 45 \degree) - \tan(A - 45 \degree) \\ \\ = \frac{ \tan(A) + \tan(45 \degree) }{1 - \tan(A) \tan(45 \degree) } - \frac{ \tan(A) - \tan(45 \degree) }{1 + \tan(A) \tan(45 \degree) } \\ \\$

but tan 45° is 1;

$= \frac{ \tan(A) }{1 - \tan(A) } - \frac{ \tan(A) }{1 + \tan(A) } \\ \\ \frac{\tan(A) + { \tan }^{3} A + \tan(A) - { \tan}^{3} A}{1 - { \tan }^{2} A } \\ \\ = (\frac{2 \tan(A) }{1 - { \tan }^{2} A}) \\ \\ = \frac{2 \sin(A) }{ \cos(A) } \times { \sec }^{2} A \\ \\ = \frac{2}{ \cos(2A) } \\ \\ = 2 \sec(2A)$

8 0

3 years ago

P(n)= 2n + 1; Find p(2+n)

inysia [295]

Substitute n with 2+n

p(2+n)=2(2+n)+1= 4+2n+1=2n+5

6 0

4 years ago

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