Question

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show that

$1 + \tan ^{2}({ \frac{2\pi}{3} }) = \sec^{2}( { \frac{2\pi}{3} }) \:$




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Answer 1

Answer:

Step-by-step explanation:

note :

tanx = sinx/cosx      and secx = 1/cosx

1+tan²(2π/3) =  1+(sin2π/3/cos2π/3)²

= 1 +( sin2π/3)²/(cos2π/3)²

=( sin2π/3)²+(cos2π/3)²/(cos2π/3)²

but :   ( sin2π/3)²+(cos2π/3)²=1

so :1+tan²(2π/3) = 1 /(cos2π/3)²

                          =( 1/cos2π/3)²

:1+tan²(2π/3) =sec²( 2π/3)