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Physics Question on the earth's magnetic field

The plane of a dip circle is set in the geographic meridian and the apparent dip is δ1\delta_1. It is then set in a vertical plane perpendicular to the geographic meridian. The apparent dip angle is δ2\delta_2. The diclination θ\theta at the plane is

A

θ=tan⁡−1(tan⁡δ1tan⁡δ2)\theta = \tan^{-1}(\tan\delta_1 \tan \delta_2)

B

θ=tan⁡−1(tan⁡δ1+tan⁡δ2)\theta = \tan^{-1}(\tan\delta_1 + \tan \delta_2)

C

θ=tan⁡−1(tan⁡δ1tan⁡δ2)\theta = \tan^{-1}\Bigg(\frac{\tan\delta_1}{\tan \delta_2}\Bigg)

D

θ=tan⁡−1(tan⁡δ1−tan⁡δ2)\theta= \tan^{-1}(\tan\delta_1 - \tan \delta_2)

Answer

θ=tan⁡−1(tan⁡δ1tan⁡δ2)\theta = \tan^{-1}\Bigg(\frac{\tan\delta_1}{\tan \delta_2}\Bigg)

Explanation

Solution

Here, tan⁡δ1=vHcos⁡θ\tan \delta_1=\frac{v}{H \cos\theta}         tan⁡δ2=vHcos⁡(90∘−θ)=vHcos⁡θ\, \, \, \, \, \, \, \, \tan \delta_2=\frac{v}{H \cos(90^\circ-\theta)}=\frac{v}{H \cos\theta}          tan⁡δ1tan⁡δ2=sin⁡θcos⁡θ=tan⁡θ\, \, \, \, \, \, \, \, \, \frac{\tan \delta_1}{\tan \delta_2}=\frac{\sin\theta}{\cos\theta}= \tan\theta or θ=tan⁡−1(tan⁡δ1tan⁡δ2)\theta= \tan^{-1}\Bigg(\frac{\tan \delta_1}{\tan \delta_2}\Bigg)