Question

Evaluate the following $\sec {{50}^{\circ }}\sin {{40}^{\circ }}+\cos {{40}^{\circ }}\csc {{50}^{\circ }}$.

Answer 1

Hint: In the above type of trigonometric question, make all the angles equal by using complementary or supplementary angle formulae so that we can further solve it. Also we will use the concept that cosecant, sine and secant, cosine are the reciprocal of each other. i.e, we will use the following relations; \[\sec \theta =\dfrac{1}{\cos \theta }\text{ and csc}\theta =\dfrac{1}{\sin \theta }.\]

Complete step-by-step solution -
In the above question we will use the formulae as given below;
\[\sec ({{90}^{\circ }}-\theta )=\csc \theta \text{ and csc}({{90}^{\circ }}-\theta )=\sec \theta .\] These are the complementary angle formula for secant and cosecant which we will use to solve the given expression.
So, using the above formula to the given expression and further solving, we will get,
\[\begin{align}
  & \Rightarrow \sec ({{90}^{\circ }}-{{40}^{\circ }})\sin {{40}^{\circ }}+\cos {{40}^{\circ }}\csc ({{90}^{\circ }}-{{40}^{\circ }}) \\
 & \Rightarrow \csc {{40}^{\circ }}\sin {{40}^{\circ }}+\sec {{40}^{^{\circ }}}\cos {{40}^{\circ }} \\
\end{align}\]
Also, we know that relations between cosecant, sine and secant, cosine are\[\csc \theta \sin \theta =1\text{ and }\sec \theta \cos \theta =1\] respectively.
So, using the above relations between cosecant, sine and secant, cosine in the above expression we will get the following;
\[\Rightarrow 1+1=2\]
Hence, the answer will be 2.
Therefore, the value of the given expression of trigonometry on solving will be equal to 2.

NOTE: Just remember the formula of complementary angle that are mentioned above and also remember the relations between cosecant, sine and secant, cosine are already mentioned above. The first approach to the above type of question is to make the angles equal.
Be careful while doing the calculation because there is a greater chance that one can make mistakes while using the complementary angle formulae in the given expression.