Question

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

Answer 1

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 \[{\text{cosec}}\theta = \dfrac{1}{{\sin \theta }}\] .
\[\cos \left( {90 - \theta } \right) = \sin \theta ,\sin \left( {90 - \theta } \right) = \cos \theta \]


Complete step-by-step answer:
Taking LHS
Given \[\sin {48^0}\sec {42^0} + \cos {48^0}{\text{cosec}}{42^0}\]
We know that,
\[\sec \theta = \dfrac{1}{{\cos \theta }}\] and \[{\text{cosec}}\theta = \dfrac{1}{{\sin \theta }}\] .
Now given equation becomes,
\[
   \Rightarrow \sin {48^0}\sec {42^0} + \cos {48^0}{\text{cosec}}{42^0} \\
   \Rightarrow \dfrac{{\sin {{48}^0}}}{{\cos {{42}^0}}} + \dfrac{{\cos {{48}^0}}}{{\sin {{42}^0}}} \\
 \]
We have,
\[\cos \left( {90 - \theta } \right) = \sin \theta ,\sin \left( {90 - \theta } \right) = \cos \theta \]
So let us use the formula for the derivation.
\[
   \Rightarrow \dfrac{{\sin {{48}^0}}}{{\cos {{\left( {90 - 48} \right)}^0}}} + \dfrac{{\cos {{48}^0}}}{{\sin {{\left( {90 - 48} \right)}^0}}} \\
   \Rightarrow \dfrac{{\sin {{48}^0}}}{{\sin {{48}^0}}} + \dfrac{{\cos {{48}^0}}}{{\cos {{48}^0}}} \\
   \Rightarrow 1 + 1 \\
   \Rightarrow 2 \\
 \]
Hence, proved \[\sin {48^0}\sec {42^0} + \cos {48^0}{\text{cosec}}{42^0} = 2\] .

Note: In this problem, you take LHS and solve it by using trigonometric formulas and prove that LHS is equal to the RHS. Students must remember different conversion formulas such as conversion from sine to cosine and other using the angle at 90 degree to get the answer. Students must not try to find the values and solve it as it will be very lengthy.