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Trigonometrical Equations
normal
$\sum\limits_{r = 1}^{100} {\frac{{\tan \,{2^{r - 1}}}}{{\cos \,{2^r}}}} $ =
A
$tan\,2^{99} -tan\,1$
B
$tan\,2^{100}$
C
$tan\,2^{100} -tan\,1$
D
એક પણ નહી
Solution
$\mathrm{T}_{\mathrm{x}}=\frac{\tan 2^{\mathrm{r}-1}}{\cos 2^{\mathrm{r}}}=\tan 2^{\mathrm{r}}-\tan 2^{\mathrm{r}-1}$
$ \Rightarrow \sum\limits_{r = 1}^{100} {{T_r} = \tan {2^{100}} – \tan 1} $
Standard 11
Mathematics
