Question 2 iShow that -tan 48- tan 23- tan 42- tan 67- - 1

Tan 48^∘ tan 23^∘ tan 42^∘ tan 67^∘ = tan (90^∘ 42^∘) tan (90^∘ 67^∘) tan 42^∘ tan 67^∘ = cot 42^∘ cot 67^∘ tan 42^∘ tan 67^∘ = (cot 42^∘ tan 42^∘) (cot 67^ ...

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Standard IX

Mathematics

Variation of Trigonometric Ratios from 0 to 90 Degrees

Question 2 iS...

Question

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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Solution

$t a n 48^{\circ} t a n 23^{\circ} t a n 42^{\circ} t a n 67^{\circ}$
$= t a n (90^{\circ} - 42^{\circ}) t a n (90^{\circ} - 67^{\circ}) t a n 42^{\circ} t a n 67^{\circ}$
$= c o t 42^{\circ} c o t 67^{\circ} t a n 42^{\circ} t a n 67^{\circ}$
$= (c o t 42^{\circ} t a n 42^{\circ}) (c o t 67^{\circ} t a n 67^{\circ})$
$= (c o t 42^{\circ} \times \frac{1}{c o t 42^{\circ}}) (c o t 67^{\circ} \times \frac{1}{c o t 67^{\circ}}) = 1 \times 1 = 1.$

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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

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

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}

= ___

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Variation of Trigonometric Ratios from 0 to 90 Degrees

Standard IX Mathematics

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Question 2 (i) Show that : tan48∘tan23∘tan42∘tan67∘=1

tan48∘tan23∘tan42∘tan67∘ =tan(90∘−42∘)tan(90∘−67∘)tan42∘tan67∘ =cot42∘cot67∘tan42∘tan67∘ =(cot42∘tan42∘)(cot67∘tan67∘) =(cot42∘×1cot42∘)(cot67∘×1cot67∘)=1×1=1.

Question 2 (i)
Show that :
$t a n 48^{\circ} t a n 23^{\circ} t a n 42^{\circ} t a n 67^{\circ} = 1$