-Show that-I-tan-48 deg- tan 23 deg-tan 42 deg- tan 67 deg- 1-II-cos-38 deg- cos 52 deg- minus- sin 38 deg- sin 52 deg- 0-

(I) tan 48 deg; tan 23 deg; tan 42 deg; tan 67 deg; = tan (90 deg; minus; 42 deg;) tan (90 deg; minus; 67 deg;) tan 42 deg; tan ...

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

Byju's Answer

Standard XII

Mathematics

Trigonometric Ratios of Compound Angles

Show that ...

Question

Show that

(I) tan 48° tan 23° tan 42° tan 67° = 1

(II)cos 38° cos 52° − sin 38° sin 52° = 0

Open in App

Solution

(I) 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° tan 67°)

= (1) (1)

= 1

(II) cos 38° cos 52° − sin 38° sin 52°

= cos (90° − 52°) cos (90°−38°) − sin 38° sin 52°

= sin 52° sin 38° − sin 38° sin 52°

= 0

Suggest Corrections

Similar questions

Q. 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

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. The value of

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

and

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

is equal to

Without using trigonometric tables, prove that:

$(i) t a n 48^{\circ} t a n 23^{\circ} t a n 42^{\circ} t a n 67^{\circ} = 1$

$(i i) t a n 7^{\circ} t a n 23^{\circ} t a n 60^{\circ} t a n 67^{\circ} t a n 83^{\circ} = \sqrt{3}$

$(i i i) c o t 12^{\circ} c o t 38^{\circ} c o t 52^{\circ} c o t 60^{\circ} c o t 78^{\circ} = \frac{1}{\sqrt{3}}$ [3 MARKS]

Join BYJU'S Learning Program

Show that (I) tan 48° tan 23° tan 42° tan 67° = 1 (II)cos 38° cos 52° − sin 38° sin 52° = 0

Show that

(I) tan 48° tan 23° tan 42° tan 67° = 1

(II)cos 38° cos 52° − sin 38° sin 52° = 0