Transform the following expression tan - - tan -cot - - cot - - -cos - - - sec - - - - 1-1-

tan α + tan βcot α +cotβ #160; + [ cos(α β )sec (α +β )+1]^ 1 = tan α +tanβtan β + tanαtanα .tanβ+ 1cos(α β )cos(α +β )+1 = tan α tan β #16 ...

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Transform the...

Question

Transform the following expression

$\frac{t a n α + t a n β}{c o t α + c o t β} + [c o s (α - β) s e c (α + β) + 1]^{- 1} .$

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Similar questions

c o t α + t a n α = m

\frac{1}{c o s α} - c o s α = n

tan \frac{α - β}{2} = \pm \sqrt{\frac{1 - cos (α - β)}{1 + cos (α - β)}}

\frac{1 + t a n α + t a n^{2} α}{1 + c o t α + c o t^{2} α} = t a n^{2} α .

cos α = \frac{3 cos β - 1}{3 - cos β}

0 < β < α < π

\frac{tan \frac{α}{2}}{t a n \frac{β}{2}} =

t a n α

=

\frac{m}{m + 1}

t a n β

=

\frac{1}{1 + 2 m}

t a n (α + β)

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