Question

(b) Without using trigonometric tables, show that:
tan 25° tan 35° tan 55° tan 65° = 1​

Answer 1

$\begin{gathered}\Large{\bold{{\underline{Formula \: Used - }}}}  \end{gathered}$

$\boxed{ \bf\:tan(90\degree\: - x) = cotx}$

$\boxed{\bf\: cotx = \dfrac{1}{tanx}}$

$\large\underline{\bf{Solution-}}$

Consider,

$\rm :\longmapsto\: tan25\degree \: tan35\degree \: tan 55\degree \: tan 65\degree \:$

$\rm :\longmapsto\: = \: (tan25\degree \: tan65\degree) \: (tan 55\degree \: tan 35\degree) \:$

$\rm :\longmapsto\: = \: (tan(90\degree \: - 65\degree) \: tan65\degree) \: (tan 55\degree \: tan (90\degree \: - 55\degree) )\:$

$\rm :\longmapsto\: = \: (cot65\degree \: tan65\degree) \: (tan 55\degree \: cot 55\degree) \:$

$\rm :\longmapsto\: = \dfrac{1}{tan65\degree \:} tan 65\degree \: tan 55\degree \:\dfrac{1}{ tan 55\degree \:}$

$\rm :\longmapsto\: = 1$

${\boxed{\boxed{\bf{Hence, Proved}}}}$

Additional Information:-

Relationship between sides and T ratios

sin θ = Opposite Side/Hypotenuse

cos θ = Adjacent Side/Hypotenuse

tan θ = Opposite Side/Adjacent Side

sec θ = Hypotenuse/Adjacent Side

cosec θ = Hypotenuse/Opposite Side

cot θ = Adjacent Side/Opposite Side

Reciprocal Identities

cosec θ = 1/sin θ

sec θ = 1/cos θ

cot θ = 1/tan θ

sin θ = 1/cosec θ

cos θ = 1/sec θ

tan θ = 1/cot θ

Co-function Identities

sin (90°−x) = cos x

cos (90°−x) = sin x

tan (90°−x) = cot x

cot (90°−x) = tan x

sec (90°−x) = cosec x

cosec (90°−x) = sec x

Fundamental Trigonometric Identities

sin²θ + cos²θ = 1

sec²θ - tan²θ = 1

cosec²θ - cot²θ = 1