Q. 2 A4.6( 18 Votes )

# Show that

We have

LHS = tan 48° tan 16° tan 42° tan 74° = (tan 48° tan 42°)(tan 16° tan 74°)

We know the trigonometric identities, we can say

tan (90° - θ) = cot θ

And tan θ = 1/cot θ

First, replace θ by 42°.

tan (90° - 42°) = cot 42°

tan 48° = cot 42° …(1)

And tan 42° = 1/cot 42° …(2)

Now, replace θ by 74°.

tan (90° - 74°) = cot 74°

tan 16° = cot 74° …(3)

And tan 74° = 1/cot 74° …(4)

Using equations (1), (2), (3) & (4), we get

LHS = tan 48° tan 16° tan 42° tan 74° = (tan 48° tan 42°)(tan 16° tan 74°)

= (cot 42°/cot 42°)(cot 74°/cot 74°)

= 1 = RHS

Thus, tan 48° tan 16° tan 42° tan 74° = 1.

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