Question

Find the value of the given: $ \sqrt{\dfrac{\sec \theta -1}{\sec \theta +1}}+\sqrt{\dfrac{\sec \theta +1}{\sec \theta -1}}=\_\_\_\_\_\_. $
A. $ 2\csc \theta $
B. $ \dfrac{2\sin \theta }{\sqrt{\sec \theta }} $
C. $ 2\cos \theta $
D. $ 2\sec \theta $

Answer 1

Hint: We know that $ {{\tan }^{2}}\theta +1={{\sec }^{2}}\theta $ .
It should also be observed that $ \tan \theta =\dfrac{\sin \theta }{\cos \theta } $ , $ \sec \theta =\dfrac{1}{\cos \theta } $ and $ \csc \theta =\dfrac{1}{\sin \theta } $ .
Equate the denominators of the expressions and simplify.

Complete step by step answer:
Let us simplify the given expression:
 $ \sqrt{\dfrac{\sec \theta -1}{\sec \theta +1}}+\sqrt{\dfrac{\sec \theta +1}{\sec \theta -1}} $
In order to equate the denominators, let's multiply and divide the first term by $ \sec \theta -1 $ and the second term by $ \sec \theta +1 $ :
= $ \sqrt{\dfrac{(\sec \theta -1)(\sec \theta -1)}{(\sec \theta +1)(\sec \theta -1)}}+\sqrt{\dfrac{(\sec \theta +1)(\sec \theta +1)}{(\sec \theta -1)(\sec \theta +1)}} $
Using the identity $ (a+b)(a-b)={{a}^{2}}-{{b}^{2}} $ , we get:
= $ \sqrt{\dfrac{{{(\sec \theta -1)}^{2}}}{{{\sec }^{2}}\theta -1}}+\sqrt{\dfrac{{{(\sec \theta +1)}^{2}}}{{{\sec }^{2}}\theta -1}} $
Now, using the fact that $ {{\tan }^{2}}\theta ={{\sec }^{2}}\theta -1 $ and taking the square root, we get:
= $ \dfrac{\sec \theta -1}{\tan \theta }+\dfrac{\sec \theta +1}{\tan \theta } $
= $ \dfrac{\sec \theta -1+\sec \theta +1}{\tan \theta } $
= $ \dfrac{2\sec \theta }{\tan \theta } $
Now, substituting $ \tan \theta =\dfrac{\sin \theta }{\cos \theta } $ and $ \sec \theta =\dfrac{1}{\cos \theta } $ , we get:
= $ 2\times \dfrac{\dfrac{1}{\cos \theta }}{\dfrac{\sin \theta }{\cos \theta }} $
= $ 2\times \dfrac{1}{\cos \theta }\times \dfrac{\cos \theta }{\sin \theta } $
= $ \dfrac{2}{\sin \theta } $
= $ 2\csc \theta $ ... (since $ \csc \theta =\dfrac{1}{\sin \theta } $ )

The correct answer is A. $ 2\csc \theta $ .

Note: In a right-angled triangle with length of the side opposite to angle θ as perpendicular (P), base (B) and hypotenuse (H):
 $ \sin \theta =\dfrac{P}{H},\cos \theta =\dfrac{B}{H},\tan \theta =\dfrac{P}{B} $
 $ {{P}^{2}}+{{B}^{2}}={{H}^{2}} $ (Pythagoras' Theorem)
The identities $ {{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1 $ , $ {{\tan }^{2}}\theta +1={{\sec }^{2}}\theta $ and $ 1+{{\cot }^{2}}\theta ={{\csc }^{2}}\theta $ are equivalent to each other and they are a direct result of the Pythagoras' theorem.