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, thenYou need to login to perform this action.
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