Question

all formula of trigonometry don't give the table

Answer 1

heyy dude here is ur answer

sin = p/h

cos =b/h

tan=p/b

cosec=h/p

sec=h/b

cot=b/p

hope it will help u but plzz mark mine answer as the brainliest one

Answer 2

${ \red{ \bf{ Information \: related \: to \:Trigonometry:}}}$

${ \green{ \bf{ sin θ = Perpendicular/Hypotenuse }}}$

${ \green{ \bf{ cos θ = Base/Hypotenuse }}}$

${ \green{ \bf{tan θ = Perpendicular/Base }}}$

${ \green{ \bf{sec θ = Hypotenuse/Base }}}$

${ \green{ \bf{ cosec θ = Hypotenuse/Perpendicular }}}$

${ \green{ \bf{ cot θ = Base/Perpendicular }}}$

${ \red{ \bf{Their \: reciprocal \: Identities: }}}$

${ \green{ \bf{ cosec θ = 1/sin θ }}}$

${ \green{ \bf{ sec θ = 1/cos θ }}}$

${ \green{ \bf{ cot θ = 1/tan θ }}}$

${ \green{ \bf{sin θ = 1/cosec θ }}}$

${ \green{ \bf{ cos θ = 1/sec θ }}}$

${ \green{ \bf{ tan θ = 1/cot θ}}}$

${ \red{ \bf{ Their \: co-function \: Identities: }}}$

${ \green{ \bf{ sin (90°−x) = cos x }}}$

${ \green{ \bf{cos (90°−x) = sin x }}}$

${ \green{ \bf{ tan (90°−x) = cot x }}}$

${ \green{ \bf{ cot (90°−x) = tan x }}}$

${ \green{ \bf{ sec (90°−x) = cosec x }}}$

${ \green{ \bf{ cosec (90°−x) = sec x }}}$

${ \red{ \bf{ Their \: fundamental \: trigonometric \: identities: }}}$

${ \green{ \bf{ sin²θ + cos²θ = 1 }}}$

${ \green{ \bf{ sec²θ - tan²θ = 1 }}}$

${ \green{ \bf{ cosec²θ - cot²θ = 1 }}}$