Question

Q=>Show that :

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

(ii) cos 38° cos 52° – sin 38° sin 52° = 0​

Answer 1

HERE IS YOUR ANSWER MATE.....;

(i) tan 48° tan 23° tan 42° tan 67°

We can also write the above given tan functions in terms of cot functions, such as;

tan 48° = tan (90° – 42°) = cot 42°

tan 23° = tan (90° – 67°) = cot 67°

Hence, substituting these values, we get

= cot 42° cot 67° tan 42° tan 67°

= (cot 42° tan 42°) (cot 67° tan 67°)

= 1 × 1 [since cot A.tan A = 1]

= 1

(ii) cos 38° cos 52° – sin 38° sin 52°

We can also write the given cos functions in terms of sin functions.

cos 38° = cos (90° – 52°) = sin 52°

cos 52°= cos (90° – 38°) = sin 38°

Hence, putting these values in the given equation, we get;

sin 52° sin 38° – sin 38° sin 52° = 0

Hope it's Helpful....:)

Answer 2

Answer:

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