Question

If $$\tan \dfrac{A}{2} =r$$ then the value of $$\left( \sec A+\tan A\right) $$ is equal to
A) $$\dfrac{2+r}{2-r}$$
B) $$\dfrac{2-r}{2+r}$$
C) $$\dfrac{1+r}{1-r}$$
D) $$\dfrac{1-r}{1+r}$$

Answer 1

Hint: In this question it is given that we have to find the value of $$\left( \sec A+\tan A\right) $$, where $$\tan \dfrac{A}{2} =r$$. So to find the solution we have to write $\sec A$, $ \tan A$ as $$\dfrac{1}{\cos A}$$, $$\dfrac{\sin A}{\cos A}$$ respectively. Also we have to change the angle $A$ to $\dfrac{A}{2}$, because as we know that $$\tan \dfrac{A}{2} =r$$ and then after simplification we will get our required solution.
Complete step-by-step solution:
Let the given expression as
$$S=\left( \sec A+\tan A\right) $$
$$=\left( \dfrac{1}{\cos A} +\dfrac{\sin A}{\cos A} \right) $$
$$=\left( \dfrac{1+\sin A}{\cos A} \right) $$
Now as we know that,
$$\cos \theta =\cos^{2} \dfrac{\theta }{2} -\sin^{2} \dfrac{\theta }{2}$$, and
$$\sin \theta =2\sin \dfrac{\theta }{2} \cos \dfrac{\theta }{2}$$
so by using this formula, we get,
$$S=\dfrac{1+2\sin \dfrac{A}{2} \cos \dfrac{A}{2} }{\cos^{2} \dfrac{A}{2} -\sin^{2} \dfrac{A}{2} }$$
$$=\dfrac{\sin^{2} \dfrac{A}{2} +\cos^{2} \dfrac{A}{2} +2\sin \dfrac{A}{2} \cos \dfrac{A}{2} }{\cos^{2} \dfrac{A}{2} -\sin^{2} \dfrac{A}{2} }$$ [ since, $$\sin^{2} \dfrac{A}{2} +\cos^{2} \dfrac{A}{2} =1$$]
As we know that,
$$a^{2}+2ab+b^{2}=\left( a+b\right)^{2} $$.......(1) and
$$a^{2}-b^{2}=\left( a+b\right) \left( a-b\right) $$.....(2),
so by applying identity (1) in numerator and identity (2) in denominator we get,
$$S=\dfrac{\left( \cos \dfrac{A}{2} +\sin \dfrac{A}{2} \right)^{2} }{\left( \cos \dfrac{A}{2} +\sin \dfrac{A}{2} \right) \left( \cos \dfrac{A}{2} -\sin \dfrac{A}{2} \right) }$$ [ taking $$a=\cos \dfrac{A}{2}$$ & $$b=\sin \dfrac{A}{2}$$]
$$=\dfrac{\left( \cos \dfrac{A}{2} +\sin \dfrac{A}{2} \right) }{\left( \cos \dfrac{A}{2} -\sin \dfrac{A}{2} \right) }$$
Now dividing numerator and denominator by $$\cos \dfrac{A}{2}$$, we get,
$$S=\dfrac{1+\left( \dfrac{\sin \dfrac{A}{2} }{\cos \dfrac{A}{2} } \right) }{1-\left( \dfrac{\sin \dfrac{A}{2} }{\cos \dfrac{A}{2} } \right) }$$
$$=\dfrac{1+\tan \dfrac{A}{2} }{1-\tan \dfrac{A}{2} }$$
$$=\dfrac{1+r}{1-r}$$ [since, $$\tan \dfrac{A}{2} =r$$]
Hence, the correct option is option C.
Note: While simplifying a big expression, try to express it in terms of one or two basic trigonometric functions, like we have transformed the above expression in terms of $cosine$ and $sine$, also try to find an order in the problem to apply trigonometric identities, properties and transformations.