Show that - i tan 48-tan 23-tan 42-tan 67- - 1 ii cos 38-cos 52- - sin 38-sin 52- - 0-

We know that #160; tan(90 θ)=θ #160;and #160; cos(90 θ)=sinθ (i) #160; tan (90 42)^o tan (90 67)^o tan 42^o tan 67^o = 42^o 67^o tan ...

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Byju's Answer

Standard X

Mathematics

Trigonometric Identities

Show that : i...

Question

Show that : (i) $tan 48^{\circ} tan 23^{\circ} tan 42^{\circ} tan 67^{\circ} = 1$ (ii) $cos 38^{\circ} cos 52^{\circ} - sin 38^{\circ} sin 52^{\circ} = 0$ .

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Solution

we know that $tan (90 - θ) = cot θ$ and $cos (90 - θ) = sin θ$

(i) $tan (90 - 42)^{o} tan (90 - 67)^{o} tan 42^{o} tan 67^{o}$
$= cot 42^{o} cot 67^{o} tan 42^{o} tan 67^{o}$
$= (1) (1) = 1$

(ii) $cos (96 - 52)^{o} cos (90 - 38)^{o} - sin 38^{o} sin 52^{o}$
$= sin 52^{o} sin 38^{o} - sin 38^{o} sin 52^{o}$
$= 0$

Hence Proved.

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

Q. Show that:

c o s 38^{\circ} c o s 52^{\circ} - s i n 38^{\circ} s i n 52^{\circ} = 0

Q. Question 2 (ii)
Show that :

c o s 38^{\circ} c o s 52^{\circ} - s i n 38^{\circ} s i n 52^{\circ} = 0

Q. show that

cos 38^{\circ} cos 52^{\circ} - sin 38^{\circ} sin 52^{\circ} = 0

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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Show that : (i) tan48∘tan23∘tan42∘tan67∘=1 (ii) cos38∘cos52∘−sin38∘sin52∘=0.

we know that tan(90−θ)=cotθ and cos(90−θ)=sinθ(i) tan(90−42)otan(90−67)otan42otan67o=cot42ocot67otan42otan67o=(1)(1)=1(ii) cos(96−52)ocos(90−38)o−sin38osin52o=sin52osin38o−sin38osin52o=0Hence Proved.

Show that : (i) $tan 48^{\circ} tan 23^{\circ} tan 42^{\circ} tan 67^{\circ} = 1$ (ii) $cos 38^{\circ} cos 52^{\circ} - sin 38^{\circ} sin 52^{\circ} = 0$ .