Question

$\huge \fbox \red{❥ Question}$

Show that :

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

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

Answer 1

Answer:

Step-by-step explanation:

ANSWER:-

Given That:-

• tan 48° tan 23° tan 42° tan 67°
• cos 38° cos 52° – sin 38° sin 52°

Concept:-

Trigonometry and its applications

Let's Do!

We need to know that:-

$\sf{tan \ \theta = cot(90 - \theta)}$

So, we can write tan 48 and tan 23 as:-

$\sf{tan \ 48 = cot(90 - 48) }$

$\implies \sf{cot \ 42}$

$\sf{tan \ 23 = cot(90-23)}$

$\mapsto \sf{cot \ 67}$

So, as they are complementary, they get cancelled, and hence the answer is 1.

Hence Proved!

Now, we need to know that:-

$\rm{sin \ \theta = cos(90- \theta)}$

So, we can write sin 38° sin 52° as :-

$\rm{sin \ 38 = cos(90- 38)}$

$\mapsto \rm{cos \ 52}$

$\rm{sin \ 52 = (90-52)}$

$\mapsto \rm{cos \ 38}$

cos 38° cos 52° – cos 38° cos 52° = 0

Hence Proved!

Answer 2

☆ ANSWER ☆:-

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

LHS = tan 48° tan 23° tan 42° tan 67° 

= tan 48° × tan 23° × tan (90 – 48).tan (90 – 23) 

= tan 48° × tan 23° × cot 48° × cot 23° = 1 

∴ LHS = RHS 

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

LHS = cos38° cos 52° – sin 38° sin 52° 

= cos 38° × cos (90 – 38) – sin 38° × sin (90 – 38) 

= cos 38° × sin 38° – sin 38° × cos 38° 

= 0.

• I Hope it's Helpful Ayan.