Question

Express $\sec {50^0} + \cot {78^0}$ in terms of t-ratios of angle between ${0^o}$ and ${45^0}$.

Answer 1

Hint : In $\sec {50^0} + \cot {78^0}$ , we observe that ${50^0}$ and ${78^0}$ are not in between ${0^o}$ and ${45^0}$. That is, angle does not lie in between ${0^o}$ and ${45^0}$. We need to express this in terms of $\sin \theta$,$\cos \theta$,$\tan \theta$, $\csc \theta$ ,$\sec \theta$ and $\cot \theta$. Such that $\theta$ lies between ${0^o}$ and ${45^0}$.

Complete step-by-step answer :
We have, $\sec {50^0} + \cot {78^0}$
We can express $\sec {50^0}$ as $\sec {50^0} = \sec ({90^0} - {40^0})$, because we know that $90 - 40 = 50$
And $\cot {78^0}$ as $\cot {78^0} = \cot ({90^0} - {12^0})$, because $90 - 12 = 78$.
After conversion always check where the angle lies in, that is in which quadrant, depending upon the quadrant the sign will change.
We converted the angles into ${90^0}$ because we know the standard values at that angle. We can also convert the angles into ${90^0} + \theta$ and ${180^0} \pm \theta$, if they give angles more than 90.
Then, $\sec {50^0} + \cot {78^0}$
Substituting the values,
$= \sec ({90^0} - {40^0}) + \cot ({90^0} - {12^0})$
$= \csc ({40^0}) + \tan ({12^0})$
Because, we know the trigonometric ratios $\sec ({90^0} - \theta ) = \csc (\theta )$ and $\cot ({90^0} - {12^0}) = \tan (\theta )$.
${90^0} - \theta$ Lies in the first quadrant so all the six trigonometric functions are positive.
Hence, the solution is $\csc ({40^0}) + \tan ({12^0})$. Thus we expressed in terms of t ratios of angle between ${0^o}$ and ${45^0}$.
We can see that $\theta$ lies between ${0^o}$ and ${45^0}$. If not again we continue the same procedure as above mentioned.
So, the correct answer is “$\csc ({40^0}) + \tan ({12^0})$”.

Note : In problems, if you get $\sec {120^0} + \cot {150^0}$, we can express the angles as $({90^0} + {30^0})$ and $({180^0} - {30^0})$ respectively. For these we have a formulas, $\sec ({90^0} + \theta ) = - \csc \theta$ and $\cot ({180^0} - \theta ) = - \cot \theta$, since both are in second quadrant hence the negative signs. Similarly we can express any angels in between ${0^o}$ and ${45^0}$. Using trigonometric ratios we can solve any.In this type of questions students always need to remember all the trigonometric ratios otherwise it will be difficult to solve these kinds of questions.