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Mathematics Question on Trigonometric Identities

If (1+tan⁡ 1∘)(1+tan⁡ 2∘)......(1+tan⁡ 45∘)=2n(1 + \tan \, 1^{\circ})(1 + \tan \, 2^{\circ}) ...... (1 + \tan \, 45^{\circ}) = 2^n, then n is

A

22

B

24

C

23

D

12

Answer

23

Explanation

Solution

We have,
(1+tan⁡1∘)(1+tan⁡2∘)….(1+tan⁡45∘)=2n\left(1+\tan 1^{\circ}\right)\left(1+\tan 2^{\circ}\right) \ldots .\left(1+\tan 45^{\circ}\right)=2^{ n }
Now, (1+tan⁡θ)(1+tan⁡(45∘−θ))(1+\tan \theta)\left(1+\tan \left(45^{\circ}-\theta\right)\right)
=(1+tan⁡θ)(1+1−tan⁡θ1+tan⁡θ)=(1+\tan \theta)\left(1+\frac{1-\tan \theta}{1+\tan \theta}\right)
=(1+tan⁡θ)(1+tan⁡θ+1−tan⁡θ)1+tan⁡θ=2=(1+\tan \theta) \frac{(1+\tan \theta+1-\tan \theta)}{1+\tan \theta}=2
∴(1+tan⁡∘)(1+tan⁡44∘)(1+tan⁡2∘)(1+tan⁡43∘)\therefore\left(1+\tan ^{\circ}\right)\left(1+\tan 44^{\circ}\right)\left(1+\tan 2^{\circ}\right)\left(1+\tan 43^{\circ}\right)
…(1+tan⁡22∘)(1+tan⁡23∘)(1+tan⁡45∘)\ldots\left(1+\tan 22^{\circ}\right)\left(1+\tan 23^{\circ}\right)\left(1+\tan 45^{\circ}\right)
=2×2×2………22=2\times2\times2\ldots\ldots\ldots 22 times ×(1+1)\times \left(1+1\right)
=222×2=223=2^{22}\times2=2^{23}
∴n=23\therefore n=23