Question

# What is cot A + cosec A equal to?$A.\cot \dfrac{A}{2} \\ B.\tan \dfrac{A}{2} \\ C.2\cot \dfrac{A}{2} \\ D.2\tan \dfrac{A}{2} \\$

Hint: In order to solve this problem we will use the formulas of trigonometry like $\cot A = \dfrac{{\cos A}}{{\sin A}}$ and $\cos ecA = \dfrac{1}{{\sin A}}$ then we will put these values in the given equation and solve. Then we have to use the formulas $\sin \theta = 2\sin \dfrac{\theta }{2}\cos \dfrac{\theta }{2}$ and $\cos \theta = 2{\cos ^2}\dfrac{\theta }{2} - 1$. Doing this will solve your problem.

We know that $\cot A = \dfrac{{\cos A}}{{\sin A}}$ and $\cos ecA = \dfrac{1}{{\sin A}}$ so, on putting these values in the above equation we get,
$\Rightarrow \dfrac{{\cos A}}{{\sin A}} + \dfrac{1}{{\sin A}}$ so, the new equation is $\dfrac{{\cos A + 1}}{{\sin A}}$.
Now on using the above equation we will use the formulas $\sin \theta = 2\sin \dfrac{\theta }{2}\cos \dfrac{\theta }{2}$ and $\cos \theta = 2{\cos ^2}\dfrac{\theta }{2} - 1$.
$\Rightarrow \dfrac{{2{{\cos }^2}\dfrac{\theta }{2} - 1 + 1}}{{2\sin \dfrac{\theta }{2}\cos \dfrac{\theta }{2}}}$
$\Rightarrow \dfrac{{\cos \dfrac{A}{2}}}{{\sin \dfrac{A}{2}}} = \cot \dfrac{A}{2}$
So, we got the value of $\cot A + \cos ecA = \cot \dfrac{A}{2}$.
Note: When you get to solve such problems of trigonometry you need to simply solve by using various formulas of trigonometry to reach the right answer. Generally students leave such problems but these can be solved by solving the equation using various identities to get the right answer. Trigonometric identities are equalities that involve trigonometric functions and are true for every value of the occurring variables where both sides of the equality are defined. Geometrically, these are identities involving certain functions of one or more angles. Some of the formulas used here to get this problem solved is $\cot A = \dfrac{{\cos A}}{{\sin A}}$ and $\cos ecA = \dfrac{1}{{\sin A}}$ then we have put these values in the given equation and solved. Then we have used the formulas $\sin \theta = 2\sin \dfrac{\theta }{2}\cos \dfrac{\theta }{2}$ and $\cos \theta = 2{\cos ^2}\dfrac{\theta }{2} - 1$. Doing this will solve such problems and will give you the right answer.