Question

Find the value of following trigonometric function:
$\dfrac{{\sin {{35}^0}}}{{\cos {{55}^0}}} + \dfrac{{\tan {{37}^0}}}{{\cot {{53}^0}}}$

Answer 1

Hint: Try to break the angle as a sum of other angles with multiple of ${90^0},{180^0},{270^0} \& {360^0}$. In order to solve this question we will use trigonometric identities such as
$\cos (90 - \theta ) = \sin \theta {\text{ and }}\cot (90 - \theta ) = \tan \theta$

Complete step-by-step answer:
Given term is: $\dfrac{{\sin {{35}^0}}}{{\cos {{55}^0}}} + \dfrac{{\tan {{37}^0}}}{{\cot {{53}^0}}}$
In order to make the term simpler, we will use the trigonometric identities to convert $cos {\theta}$ into $sin{\theta}$ and $cot{\theta}$ into $tan{\theta}$
As we know the trigonometric identities
$\cos (90 - \theta ) = \sin \theta {\text{ and }}\cot (90 - \theta ) = \tan \theta$
So we use the same in the above term
$\Rightarrow \dfrac{{\sin {{35}^0}}}{{\cos {{55}^0}}} + \dfrac{{\tan {{37}^0}}}{{\cot {{53}^0}}} \\\ \Rightarrow \dfrac{{\sin {{35}^0}}}{{\cos \left( {{{90}^0} - {{35}^0}} \right)}} + \dfrac{{\tan {{37}^0}}}{{\cot \left( {{{90}^0} - {{37}^0}} \right)}} \\\ \Rightarrow \dfrac{{\sin {{35}^0}}}{{\sin {{35}^0}}} + \dfrac{{\tan {{37}^0}}}{{\tan {{37}^0}}} \\\$
Now let us simplify the terms by mere cancellation.
$\Rightarrow \dfrac{{\sin {{35}^0}}}{{\sin {{35}^0}}} + \dfrac{{\tan {{37}^0}}}{{\tan {{37}^0}}} \\\ = 1 + 1 \\\ = 2 \\\$
Hence, the value of given term $\dfrac{{\sin {{35}^0}}}{{\cos {{55}^0}}} + \dfrac{{\tan {{37}^0}}}{{\cot {{53}^0}}}$ is 2.

Note: To solve this question, we used the trigonometric identities and some manipulation. Whenever we have an unknown or random angle in the problem, whose trigonometric values are unknown, try to manipulate some angle by using trigonometric identities in order to cancel that term or to bring the angle in some known value. Remember the trigonometric identities.