Question

tan 10º tan 20° tan 40° tan 50° tan 70° tan 80°

NOTE:- EVALUATE WITHOUT USING TABLE.​

Answer 1

$\large\underline\blue{\bold{Given \: Question :- }}$

Evaluate : tan 10º tan 20° tan 40° tan 50° tan 70° tan 80°

─━─━─━─━─━─━─━─━─━─━─━─━─

$\bf \:\large \red{AηsωeR : } ✍$

─━─━─━─━─━─━─━─━─━─━─━─━─

$\large\underline\blue{\bold{Formula \: used:- }}$

$\bf \:(1). \: tan(90° - x) = cotx$

$\bf \:(2). \: cotx = \dfrac{1}{tanx}$

─━─━─━─━─━─━─━─━─━─━─━─━─

$\large\underline\purple{\bold{Understanding \: the \: Concept :- }}$

Here the concept of Trigonometric Identities of complementary angles are used. In this question mainly we will use 2 identities. Firstly, we pair the terms whom sum of the angle is 90°, using first identity, we shall simplify the term in to same angle form. Then using the second identity we will find the the main equation to be solved. We will keep on simplifying and then finally we will get our answer.

Let's do it now!!

─━─━─━─━─━─━─━─━─━─━─━─━─

$\large\underline\purple{\bold{Solution :- }}$

$\bf \:  tan 10º tan 20° tan 40° tan 50° tan 70° tan 80°$

$\sf \: = (tan 10º tan 80°) (tan 40° tan 50° )(tan 20° tan 70°)$

$\sf \sf \: = \: (tan 10º \times tan (90° - 10°) )(tan 40° \times \\ \sf \: tan (90° - 40°))( tan 70° \times tan (90° - 70°))$

$\sf \: = (tan 10º cot 10°)( tan 40° cot 40° )(tan 70° cot \: 70°)$

$\sf \: = (tan10°\dfrac{1}{tan10°} )(tan40°\dfrac{1}{tan40°} )(tan70°\dfrac{1}{tan70°} )$

$\sf \: = 1$

─━─━─━─━─━─━─━─━─━─━─━─━─

$\large \green{\bf \: ⟼ Explore \: more } ✍$

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

$\begin{gathered}\begin{gathered}\begin{gathered}\begin{gathered}\bf Trigonometry\: Table \\ \begin{gathered}\begin{gathered}\begin{gathered}\begin{gathered}\boxed{\boxed{\begin{array}{ |c |c|c|c|c|c|} \bf\angle A & \bf{0}^{ \circ} & \bf{30}^{ \circ} & \bf{45}^{ \circ} & \bf{60}^{ \circ} & \bf{90}^{ \circ} \\ \\ \rm sin A & 0 & \dfrac{1}{2}& \dfrac{1}{ \sqrt{2} } & \dfrac{ \sqrt{3}}{2} &1 \\ \\ \rm cos \: A & 1 & \dfrac{ \sqrt{3} }{2}& \dfrac{1}{ \sqrt{2} } & \dfrac{1}{2} &0 \\ \\ \rm tan A & 0 & \dfrac{1}{ \sqrt{3} }&1 & \sqrt{3} & \rm \infty \\ \\ \rm cosec A & \rm \infty & 2& \sqrt{2} & \dfrac{2}{ \sqrt{3} } &1 \\ \\ \rm sec A & 1 & \dfrac{2}{ \sqrt{3} }& \sqrt{2} & 2 & \rm \infty \\ \\ \rm cot A & \rm \infty & \sqrt{3} & 1 & \dfrac{1}{ \sqrt{3} } & 0\end{array}}}\end{gathered}\end{gathered}\end{gathered} \end{gathered}\end{gathered}\end{gathered}\end{gathered}\end{gathered}$

Answer 2

Answer:

1

Step-by-step explanation:

Mark me as Brainliest.