floor equation.

\hspace{-16}$Find Sum of $\bf{\sec(40^{0})+\sec(80^{0})+\sec{160^0}=}$

2 Answers

1057
Ketan Chandak ·

1cos 40+1cos 80+1cos 160

=1cos 40+1cos 80-1cos 20

=cos 40+cos 80cos 40 x cos 80-1cos 20

=2cos 60 x cos 20cos 40 x cos 80-1cos 20

=cos220- (cos 40 x cos 80)cos 40 x cos 80 x cos 20

=cos220- (cos(60-20) x cos (60+20))cos 40 x cos 80 x cos 20

now using cos(a+b)cos(a-b)=cos2b=sin2a

we get sin260cos 40 x cos 80 x cos 20

now we can easily prove that cos 40 x cos 80 x cos 20 is equal to 18 by using the trick of multiplying sin 20 in the numerator and denominator......

thus we get the given expression sec 40+sec 80+sec 160 equal to 6....

1708
man111 singh ·

Yes Ketan It is = 6

Thanks for Nice post...

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