Question

Show that :-

$\tan48 {}^{0} \tan23 {}^{0} \tan42 {}^{0} \tan67 {}^{0} = 1$
Thanks

Answer 1

Hi...

=> tan48° tan23° tan 42° tan67°

=> tan48° tan42° tan23° tan67°

=> cot(90-48) tan (42) cot(90-23) tan(67)

=> cot42° tan42° cot67° tan67°


[cot Φ =1/tanΦ]


=> 1 × 1

=> 1


Hope it hlpz..

Answer 2

Heya !!

Here's your answer.. ⬇⬇
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$= \tan(48) \times \tan(23) \times \tan(42) \times \tan(67) \\ \\ = \tan(48) \times \tan(42) \times \tan(23) \times \tan(67) \\ \\ = \cot(90 - 48) \times \tan(42) \times \cot(90 - 23) \times \tan(67) \\ \\ = \cot(42) \times \tan(42) \times \cot(67) \times \tan(67) \\ \\ = 1 \times 1 \\ \\ = 1$
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Identities used are :-

Cot θ × tan θ = 1
tan θ = cos ( 90-θ )

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Hope it helps..
Thanks :)