Fill in the blank-tan 48- - tan 23- - tan 42-tan 67-

Tan48^∘.tan23^∘.tan42^∘.tan67^∘ We know that cot(90^∘ θ) = tanθ ∴ tan48^∘.tan23^∘.tan42^∘.tan67^∘ = tan48^∘.tan23^∘.cot(90^∘ 42^∘).tan(90^∘ 67^∘) = tan48^ ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard X

Mathematics

Introduction to Trigonometry

Fill in the b...

Question

Fill in the blank:
$t a n 48^{\circ} . t a n 23^{\circ} . t a n 42^{\circ} . t a n 67^{\circ}$ = ___

Open in App

Solution

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

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

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

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

Suggest Corrections

Similar questions

Fill in the blanks:
i) $t a n 48^{\circ} . t a n 23^{\circ} . t a n 42^{\circ} . t a n 67^{\circ}$ = ___
ii) $c o s 38^{\circ} . c o s 52^{\circ} - s i n 38^{\circ} . s i n 52^{\circ}$ = ___

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

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

tan 48^{\circ} . tan 23^{\circ} . tan 42^{\circ} . tan 67^{\circ}

= ___

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

Join BYJU'S Learning Program

Fill in the blank: tan 48∘.tan 23∘.tan 42∘.tan 67∘ = ___

Fill in the blank:
$t a n 48^{\circ} . t a n 23^{\circ} . t a n 42^{\circ} . t a n 67^{\circ}$ = ___