Tan 48-tan 23-tan 42-tan 67-

We know that cot(90^∘ θ) = tanθ ∴ tan48^∘.tan23^∘.tan42^∘.tan67^∘ = tan48^∘.tan23^∘.cot(90^∘ 42^∘).cot(90^∘ 67^∘) = tan48^∘.tan23^∘.cot48^∘.cot23^∘ We also ...

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Byju's Answer

Standard X

Mathematics

Complementary Trigonometric ratios

tan 48∘·tan 2...

Question

$tan 48^{\circ} . tan 23^{\circ} . tan 42^{\circ} . tan 67^{\circ}$ = ___

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Solution

We know that $c o t (90^{\circ} - θ) = t a n θ$

$∴ t a n 48^{\circ} . t a n 23^{\circ} . t a n 42^{\circ} . t a n 67^{\circ}$
= $t a n 48^{\circ} . t a n 23^{\circ} . c o t (90^{\circ} - 42^{\circ}) . c o t (90^{\circ} - 67^{\circ})$
= $t a n 48^{\circ} . t a n 23^{\circ} . c o t 48^{\circ} . c o t 23^{\circ}$

We also know that $t a n θ \times c o t θ = 1$

$∴ t a n 48^{\circ} . t a n 23^{\circ} . c o t 48^{\circ} . c o t 23^{\circ}$
= $t a n 48^{\circ} . c o t 48^{\circ} . t a n 23^{\circ} . c o t 23^{\circ}$
= $1 \times 1$
= 1
Thus, $t a n 48^{\circ} . t a n 23^{\circ} . t a n 42^{\circ} . t a n 67^{\circ} = 1$

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Similar questions

t a n 48^{\circ} t a n 23^{\circ} t a n 42^{\circ} t a n 67^{\circ} =

Show that $tan 48^{\circ} tan 23^{\circ} tan 42^{\circ} tan 67^{\circ} = 1$

Q. Show that :

t a n 48^{\circ} t a n 23^{\circ} t a n 42^{\circ} t a n 67^{\circ} = 1

The value of
$t a n 48^{\circ} . t a n 23^{\circ} . t a n 42^{\circ} . t a n 67^{\circ}$ =___

Q. Question 2 (i)
Show that :

t a n 48^{\circ} t a n 23^{\circ} t a n 42^{\circ} t a n 67^{\circ} = 1

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tan 48∘.tan 23∘.tan 42∘.tan 67∘ = ___

$tan 48^{\circ} . tan 23^{\circ} . tan 42^{\circ} . tan 67^{\circ}$ = ___