Question

Show that :
(i) tan 48° tan 23° tan 42° tan 67° = 1
(ii) cos 38° cos 52° - sin 38° sin 52° = 0​

Answer 1

Explanation:

1) tan48 .tan23.tan42.tan67=1

can be written as :-

tan48.tan42.tan23.tan67=1

we know that tan

$\tan( \alpha ) = cot(90 - \alpha )$

so, tan48.cot48.tan23.cot23=1

* tan A × cot A=1*

so

1=1

LHS=RHS

Answer 2

Answer:

(i) Taking LHS

tan 48° tan 42° tan 23° tan 67°

= tan 48° tan(90°- 48°) tan 23° tan(90° -23°)

= tan 48° cot 48° tan 23° cot 23°

= (tan 48⁰ x 1/tan 48°) x (tan 23° x 1/tan 23°)

= 1 X 1

= 1

= RHS

-

(ii) LHS

= cos 38° cos 52°sin 38°. sin 52°

= cos (90°-52°). cos 52°-sin (90°52°). sin 52°

= sin 52°. cos 52° cos 52°. sin 52°

= 0 = RHS