A) \[{{\cos }^{2}}\varphi ={{\cos }^{2}}{{\varphi }_{1}}+{{\cos }^{2}}{{\varphi }_{2}}\]
B) \[{{\sec }^{2}}\varphi ={{\sec }^{2}}{{\varphi }_{1}}+{{\sec }^{2}}{{\varphi }_{2}}\]
C) \[{{\tan }^{2}}\varphi ={{\tan }^{2}}{{\varphi }_{1}}+{{\tan }^{2}}{{\varphi }_{2}}\]
D) \[{{\cot }^{2}}\varphi ={{\cot }^{2}}{{\varphi }_{1}}+{{\cot }^{2}}{{\varphi }_{2}}\]
Correct Answer: D
Solution :
Let a be the angle which one of the planes make with the magnetic meridian the other plane makes an angle \[({{90}^{o}}-\alpha )\]with it. The components of H in these planes will be \[H\cos \alpha \]and \[H\sin \alpha \] respectively. If \[{{\varphi }_{1}}\]and \[{{\varphi }_{2}}\]are the apparent dips in these two planes, then \[\tan {{\varphi }_{1}}=\frac{V}{H\cos \alpha }\] i.e. \[\cos \alpha =\frac{V}{H\tan {{\varphi }_{1}}}\] ..... (i) \[\tan {{\varphi }_{2}}=\frac{V}{H\sin \alpha }\] i.e. \[\sin \alpha =\frac{V}{H\tan {{\varphi }_{2}}}\] ..... (ii) Squaring and adding (i) and (ii), we get \[{{\cos }^{2}}\alpha +{{\sin }^{2}}\alpha ={{\left( \frac{V}{H} \right)}^{2}}\left( \frac{1}{{{\tan }^{2}}{{\varphi }_{1}}}+\frac{1}{{{\tan }^{2}}{{\varphi }_{2}}} \right)\] i.e. \[1=\frac{{{V}^{2}}}{{{H}^{2}}}\left( {{\cot }^{2}}{{\varphi }_{1}}+{{\cot }^{2}}{{\varphi }_{2}} \right)\] or \[\frac{{{H}^{2}}}{{{V}^{2}}}={{\cot }^{2}}{{\varphi }_{1}}+{{\cot }^{2}}{{\varphi }_{2}}\]i.e. \[{{\cot }^{2}}\varphi ={{\cot }^{2}}{{\varphi }_{1}}+{{\cot }^{2}}{{\varphi }_{2}}\] This is the required result.You need to login to perform this action.
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