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Question: Find the value of the given: \( \sqrt{\dfrac{\sec \theta -1}{\sec \theta +1}}+\sqrt{\dfrac{\sec \the...

Find the value of the given: \sqrt{\dfrac{\sec \theta -1}{\sec \theta +1}}+\sqrt{\dfrac{\sec \theta +1}{\sec \theta -1}}=\\_\\_\\_\\_\\_\\_.
A. 2csc⁡θ2\csc \theta
B. 2sin⁡θsec⁡θ\dfrac{2\sin \theta }{\sqrt{\sec \theta }}
C. 2cos⁡θ2\cos \theta
D. 2sec⁡θ2\sec \theta

Explanation

Solution

We know that tan⁡2θ+1=sec⁡2θ{{\tan }^{2}}\theta +1={{\sec }^{2}}\theta .
It should also be observed that tan⁡θ=sin⁡θcos⁡θ\tan \theta =\dfrac{\sin \theta }{\cos \theta } , sec⁡θ=1cos⁡θ\sec \theta =\dfrac{1}{\cos \theta } and csc⁡θ=1sin⁡θ\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:
sec⁡θ−1sec⁡θ+1+sec⁡θ+1sec⁡θ−1\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⁡θ−1\sec \theta -1 and the second term by sec⁡θ+1\sec \theta +1 :
= (sec⁡θ−1)(sec⁡θ−1)(sec⁡θ+1)(sec⁡θ−1)+(sec⁡θ+1)(sec⁡θ+1)(sec⁡θ−1)(sec⁡θ+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)=a2−b2(a+b)(a-b)={{a}^{2}}-{{b}^{2}} , we get:
= (sec⁡θ−1)2sec⁡2θ−1+(sec⁡θ+1)2sec⁡2θ−1\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⁡2θ=sec⁡2θ−1{{\tan }^{2}}\theta ={{\sec }^{2}}\theta -1 and taking the square root, we get:
= sec⁡θ−1tan⁡θ+sec⁡θ+1tan⁡θ\dfrac{\sec \theta -1}{\tan \theta }+\dfrac{\sec \theta +1}{\tan \theta }
= sec⁡θ−1+sec⁡θ+1tan⁡θ\dfrac{\sec \theta -1+\sec \theta +1}{\tan \theta }
= 2sec⁡θtan⁡θ\dfrac{2\sec \theta }{\tan \theta }
Now, substituting tan⁡θ=sin⁡θcos⁡θ\tan \theta =\dfrac{\sin \theta }{\cos \theta } and sec⁡θ=1cos⁡θ\sec \theta =\dfrac{1}{\cos \theta } , we get:
= 2×1cos⁡θsin⁡θcos⁡θ2\times \dfrac{\dfrac{1}{\cos \theta }}{\dfrac{\sin \theta }{\cos \theta }}
= 2×1cos⁡θ×cos⁡θsin⁡θ2\times \dfrac{1}{\cos \theta }\times \dfrac{\cos \theta }{\sin \theta }
= 2sin⁡θ\dfrac{2}{\sin \theta }
= 2csc⁡θ2\csc \theta ... (since csc⁡θ=1sin⁡θ\csc \theta =\dfrac{1}{\sin \theta } )

The correct answer is A. 2csc⁡θ2\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⁡θ=PH,cos⁡θ=BH,tan⁡θ=PB\sin \theta =\dfrac{P}{H},\cos \theta =\dfrac{B}{H},\tan \theta =\dfrac{P}{B}
P2+B2=H2{{P}^{2}}+{{B}^{2}}={{H}^{2}} (Pythagoras' Theorem)
The identities sin⁡2θ+cos⁡2θ=1{{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1 , tan⁡2θ+1=sec⁡2θ{{\tan }^{2}}\theta +1={{\sec }^{2}}\theta and 1+cot⁡2θ=csc⁡2θ1+{{\cot }^{2}}\theta ={{\csc }^{2}}\theta are equivalent to each other and they are a direct result of the Pythagoras' theorem.