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# Prove that $\sin {48^0}\sec {42^0} + \cos {48^0}{\text{cosec}}{42^0} = 2$ .

Last updated date: 25th Jun 2024
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Hint: In this type of problem, you take one side and solve it and get another side. Use a different conversion formula from the trigonometric identities in order to replace the different angles into one and finally cancel out the terms. Use the below formula to get the answer.
$\sec \theta = \dfrac{1}{{\cos \theta }}$ and ${\text{cosec}}\theta = \dfrac{1}{{\sin \theta }}$ .
$\cos \left( {90 - \theta } \right) = \sin \theta ,\sin \left( {90 - \theta } \right) = \cos \theta$

Given $\sin {48^0}\sec {42^0} + \cos {48^0}{\text{cosec}}{42^0}$
$\sec \theta = \dfrac{1}{{\cos \theta }}$ and ${\text{cosec}}\theta = \dfrac{1}{{\sin \theta }}$ .
$\Rightarrow \sin {48^0}\sec {42^0} + \cos {48^0}{\text{cosec}}{42^0} \\ \Rightarrow \dfrac{{\sin {{48}^0}}}{{\cos {{42}^0}}} + \dfrac{{\cos {{48}^0}}}{{\sin {{42}^0}}} \\$
$\cos \left( {90 - \theta } \right) = \sin \theta ,\sin \left( {90 - \theta } \right) = \cos \theta$
$\Rightarrow \dfrac{{\sin {{48}^0}}}{{\cos {{\left( {90 - 48} \right)}^0}}} + \dfrac{{\cos {{48}^0}}}{{\sin {{\left( {90 - 48} \right)}^0}}} \\ \Rightarrow \dfrac{{\sin {{48}^0}}}{{\sin {{48}^0}}} + \dfrac{{\cos {{48}^0}}}{{\cos {{48}^0}}} \\ \Rightarrow 1 + 1 \\ \Rightarrow 2 \\$
Hence, proved $\sin {48^0}\sec {42^0} + \cos {48^0}{\text{cosec}}{42^0} = 2$ .