Question

show that tan 48 degrees tan 16 degrees tan 42 degrees tan 74 degrees = 1​

Answer 1

To Prove:

$\tan(48) \tan(16) \tan(42) \tan(74) = 1$

Step-by-step explanation:

We have,

L.H.S

$= \tan(48) \tan(16) \tan(42) \tan(74) \\ \\ = (\tan48 \times \tan42)( \tan16 \times \tan74) \\ \\ = ( \tan48 \times \cot(90 - 42))( \tan16 \times \cot(90 - 74) ) \\ \\ = ( \tan48 \times \cot48) ( \tan16 \times \cot16) \\ \\ = 1 \times 1 \\ \\ = 1$

= R.H.S

°.° L.H.S = R.H.S

Hence, Proved.

Concept Map:-

• $\tan \alpha = \cot(90 - \alpha )$
• $\tan \alpha \times \cot \alpha = 1$

Answer 2

R.T.P :-

tan(48) tan(16) tan(42) tan(74) = 1

NOTE :-

• tan(90-a) = cot a
• tan a × cot a = 1

PROOF :-

Consider tan(48) tan(16) tan(42) tan(74) [ L.H.S]

=> tan (48) . tan (90-74) . tan (90-48) . tan(74)

=> tan (48) . cot(74). cot(48). tan(74)

Grouping the terms,

=> { tan (48). cot(48) } × { tan (74) .cot(74)}

=> 1 × 1

= 1

= R.H.S

THEREFORE L.H.S = R.H.S