I tan 48-tan 23-tan 42-tan 67-

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

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Byju's Answer

Standard X

Mathematics

Introduction to Trigonometry

i tan 48∘.tan...

Question

i) $t a n 48^{\circ} . t a n 23^{\circ} . t a n 42^{\circ} . t a n 67^{\circ}$ = ___

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Solution

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

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

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

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

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Q. i)

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

= ___

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) $c o s 38^{\circ} . c o s 52^{\circ} - s i n 38^{\circ} . s i n 52^{\circ}$ = ___