Q2 of 46 Page 288

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A. tan 48° tan16° tan 42° tan 74° = 1

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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